CN102737361B - Method for transforming full distance of three-dimensional binary image - Google Patents

Abstract

An embodiment of the present invention provides a method for complete distance transformation of a three-dimensional binary image. The method includes: dividing the three-dimensional binary image into a corresponding number of two-dimensional binary pictures along one axis; Two-dimensional complete distance transformation of each pixel in a two-dimensional binary image; determining the first closest background pixel range of the first pixel; determining each two-dimensional distance of the first pixel in the first closest background pixel range The two-dimensional complete distance transformation value of the projection pixel in the binary image; according to the two-dimensional complete distance transformation value of the first pixel in the first background pixel in the projection pixel of each two-dimensional binary image, determine the For the first background pixel closest to the first background pixel in the three-dimensional binary image, calculate a first complete distance between the first background pixel and the first pixel. The calculation speed and accuracy can be improved by adopting this embodiment.

Description

A Method for Complete Distance Transformation of 3D Binary Image

technical field

The invention relates to the field of image processing, and more specifically relates to a method for complete distance transformation of a three-dimensional binary image.

Background technique

Distance transformation is an important preprocessing operation for digital images. The digital images after distance transformation have good rotation, translation and scale invariance, so it plays an important role in various fields of image processing technology.

Performing distance transformation on a binary image refers to the process of transforming the pixel value of each pixel in the binary image into the distance between the nearest background pixel and the pixel. The pixels in the binary image can be divided into background pixels (pixel value 0) and target pixels (pixel value 1), and the distance transformation value of the background pixels is zero.

The approximate distance transformation method and the complete distance transformation method can be used to perform distance transformation on binary images. The basic idea of the approximate distance transformation method is to use an approximate template operation commonly used in image processing to calculate the movement from outside the graph to a certain point in the graph. The distance value marked in the template is usually the rounded approximate value of the Euclidean distance, and the template cannot always move along the normal direction of the boundary contour, so this kind of method must have errors, such as city block distance, chessboard distance, chamfer distance, etc.

The complete distance transformation is to calculate the exact Euclidean distance between each pixel and its nearest background pixel. The traditional method calculates the distance between each pixel and all background pixels separately, resulting in a long calculation time. The complete distance transformation method includes: boundary propagation method and dimensionality reduction method.

In summary, the existing distance transformation methods have the following disadvantages: first, the approximation method has errors and is not suitable for processing images with strict precision requirements, such as medical images; second, the traditional complete distance transformation method takes a long time to run.

Contents of the invention

In view of this, the present invention provides a method and device for complete distance transformation of a three-dimensional binary image, to overcome the error in the method of calculating distance transformation in the prior art, which is not suitable for processing images with strict precision requirements and the calculation time long question.

To achieve the above object, the present invention provides the following technical solutions:

A method for complete distance transformation of a three-dimensional binary image, the size of the three-dimensional binary image is m × n × s, comprising:

dividing the three-dimensional binary image into a corresponding number of two-dimensional binary images along one of the axes;

Calculating the two-dimensional complete distance transform of each pixel in each segmented two-dimensional binary image;

According to the complete distance between the second target pixel and the first pixel for which three-dimensional complete distance transformation needs to be calculated, and the three-dimensional complete distance transformation value of the second target pixel in the three-dimensional binary image, determine the first pixel of the first pixel. a nearest background pixel range;

determining the two-dimensional complete distance transform value of the projected pixel of the first pixel in each two-dimensional binary image within the range of the first nearest background pixel;

According to the determined two-dimensional complete distance transformation value of the first pixel in the range of the first closest background pixel of each projected pixel of the two-dimensional binary image, determine the distance from the first pixel in the three-dimensional binary image The nearest first background pixel, calculating a first complete distance between the first background pixel and the first pixel, and using the first complete distance as a three-dimensional complete distance transformation value of the first pixel.

Preferably, the three-dimensional complete distance transformation value of the second target pixel in the three-dimensional binary image is r ₁ , according to the complete distance between the second target pixel and the first pixel, and the second target pixel in The three-dimensional complete distance transformation value in the three-dimensional binary image, determining the first nearest background pixel range of the first pixel specifically includes:

determining a full distance r ₂ of the first pixel to the second target pixel;

The first closest background pixel range is the area surrounded by a circumscribed cube of a ball O ₁ with the first pixel as the center and r ₁ +r ₂ as the radius; or, the first closest background pixel The range is the area enclosed by the inscribed cube of the ball _O2 and the circumscribed cube of the ball _{O1 with the second target pixel as the center and r1 as} the radius; or, the first closest background The pixel range is the area surrounded by the inscribed cube of the ball O ₃ and the circumscribed cube of the ball O ₁ with the first pixel as the center and |r ₁ -r ₂ | as the radius.

Preferably, the three-dimensional binary image is divided into s two-dimensional binary images along the z-axis, and the first nearest background pixel range is a sphere O with the first pixel as the center and r ₁ +r ₂ as the radius The area enclosed by the circumscribed cube of ₁ , the two-dimensional binary image where the first pixel is located is z ₀ , and the determination of each two-dimensional binary image of the first pixel in the range of the first closest background pixel The two-dimensional complete distance transform values of the projected pixels in the value image specifically include:

Taking the two-dimensional binary picture _z0 as the center of symmetry and sequentially determining the two-dimensional complete distance transformation value of the projected pixel of the first pixel in the two-dimensional binary picture _z0 + _rx from the inside to the outside along the z-axis direction The square of DR _x ² , and record the square of the determined two-dimensional complete distance transformation value, r _x =0,±1,±2,…,±(r ₁ +r ₂ );

Determine whether DR _x ² is satisfied If yes, stop determining the two-dimensional complete distance transformation of the projected pixels of the first pixel in the two-dimensional binary image z ₀ ±(r _x +1) to the two-dimensional binary image z ₀ ±(r ₁ +r ₂ ) value, if not, continue to determine the two-dimensional complete distance transformation value of the projected pixel of the first pixel in the two-dimensional binary image z ₀ ±(r x +1), until the first pixel is determined to be in the two-dimensional binary image z 0 ±(r _x +1) up to the two-dimensional complete distance transformed value of the projected pixel in the value picture z ₀ ±(r ₁ +r ₂ ).

Preferably, according to the two-dimensional complete distance transformation value of the projected pixel of each two-dimensional binary image of the first pixel in the range of the first closest background pixel, the distance in the three-dimensional binary image is determined The nearest first background pixel of the first pixel specifically includes:

Using the formula min{DR _x ² +|z _x -z ₀ | ² }, where z _x ∈ the value range of z in the first nearest background pixel range, determine the first background pixel closest to the first pixel;

use the formula Calculate the first complete distance.

Preferably, calculating the two-dimensional complete distance transformation of each two-dimensional binary image specifically includes:

A preprocessing step, the preprocessing step is to determine the first function or determine the second function, the first function is used to determine the position of the background pixel closest to the fourth pixel in the i-th row of the two-dimensional binary image, the The second function is used to determine the position of the background pixel closest to the fourth pixel in the jth column of the two-dimensional binary image, wherein, 1≤i≤m, 1≤j≤n, i and j are both integers;

According to the complete distance between the third target pixel and the fourth pixel for which the two-dimensional complete distance needs to be calculated, and the two-dimensional complete distance of the third target pixel in the two-dimensional binary image, determine the first position of the fourth pixel 2. The closest background pixel range, the fourth pixel and the third target pixel are located in the same two-dimensional binary image;

Use the first function to search for the background pixel closest to the fourth pixel in each row of the second closest background pixel range, or use the second function to search for the background pixel in the second closest background pixel range Search for the nearest background pixel to the fourth pixel in each column;

Determine the second background pixel closest to the fourth pixel from the searched background pixels closest to the fourth pixel, and calculate the second two-dimensional completeness between the second background pixel and the fourth pixel distance, and use the second two-dimensional complete distance as the two-dimensional complete distance transformation value of the fourth pixel in the two-dimensional binary picture where it is located.

Preferably, calculating the two-dimensional complete distance transformation of each two-dimensional binary image specifically includes:

A preprocessing step, the preprocessing step is to determine the first function or determine the second function, the first function is used to determine the position of the background pixel closest to the fourth pixel in the i-th row of the two-dimensional binary image, the The second function is used to determine the position of the background pixel closest to the fourth pixel in the jth column of the two-dimensional binary image, wherein, 1≤i≤m, 1≤j≤n, i and j are both integers;

According to the complete distance between the third target pixel and the fourth pixel, and the distance between the third target pixel and its nearest background pixel, determine the second closest background pixel range of the fourth pixel, and divide the second closest background pixel range according to The preset rule is divided into a first sub-nearest background pixel range set and a second sub-nearest background pixel range set, the first sub-nearest background pixel range set includes at least one first sub-nearest background pixel range, and the at least one sub-nearest background pixel range The number of rows in the range is not greater than the number of columns, the second sub-nearest background pixel range set includes at least one second sub-nearest background pixel range, and the number of rows in the at least one second sub-nearest background pixel range is greater than the number of columns;

Use the first function to search for the background pixel closest to the fourth pixel in each row of the first sub-nearest background pixel range set, and use the second function to search for the closest background pixel to the fourth pixel in each column of the second sub-nearest background pixel range set background pixels;

Determine the second background pixel closest to the fourth pixel from the searched background pixels closest to the fourth pixel, calculate the second complete distance between the second background pixel and the fourth pixel, and use this second complete distance as The 2D full distance transformed value of the fourth pixel.

Preferably, the determining the second nearest background pixel range of the fourth pixel specifically includes:

Calculating the two-dimensional complete distance r ₃ of the third target pixel in the two-dimensional binary image;

determining a complete distance r ₄ of the fourth pixel from the third target pixel;

The second closest background pixel range is the area surrounded by the circumscribed square of the circle O ₄ with the fourth pixel as the center and r ₃ +r ₄ as the radius;

Or, the second closest background pixel range is an annular area surrounded by the inscribed square of the circle _O5 and the circumscribed square of the circle _O4 with the third target pixel as the center and _r3 as the radius ;

Or, the range of the second closest background pixel is defined by the inscribed square of the circle O ₆ with the fourth pixel as the center and the radius |r ₃ -r ₄ | and the circumscribed square of the circle O ₄ enclosed circular area.

Preferably, the fourth pixel is located in the xth row and the yth column of the two-dimensional binary picture, and (x, y) represents the position of the fourth pixel in the two-dimensional binary picture, and (x ,y) is the demarcation point, which divides row x into left and right, and divides column y into top and bottom;

The determining the first function specifically includes:

Determining the first sub-function used to calculate the column number of the background pixel closest to the fourth pixel (x, y) on the left side of the xth row of the two-dimensional binary image, and used to calculate the two-dimensional binary image The second sub-function of the column number of the background pixel closest to the fourth pixel (x, y) on the right side of the xth row of the picture, wherein, 1≤x≤m, 1≤y≤n;

determining the first function according to the first sub-function and the second sub-function;

The determining the second function specifically includes:

Determine the third sub-function used to calculate the row number of the background pixel closest to the fourth pixel above the yth column of the two-dimensional binary image, and to calculate the distance below the yth column of the two-dimensional binary image The fourth sub-function of the row number of the nearest background pixel of the fourth pixel;

The second function is determined according to the third sub-function and the fourth sub-function.

Preferably, the first sub-function is specifically L[x, y]:

$\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; the\; y</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and y=1, then L(x, y)=-Maxlable;

The second sub-function is specifically R[x, y]:

$\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; the\; y</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and y=n, then R[x, y]=Maxlable;

Then the first function determined according to the first sub-function and the second sub-function is specifically SZ[x, y]:

$\mathrm{<span\; class="notranslate">\; SZ</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; <</span>\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}\\ \mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

The third sub-function is specifically T[x, y]:

$\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; x</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and x=1, then T[x, y]=-Maxlable;

The fourth sub-function is specifically D[x, y]:

$\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; x</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and x=m, then D[x, y]=Maxlable;

Then the second function determined according to the third sub-function and the fourth sub-function is specifically ZS[x, y]:

$\mathrm{<span\; class="notranslate">\; ZS</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; <</span>\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; x</span>}\\ \mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; x</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

Wherein, I(x, y)=1 represents a target pixel, I(x, y)=0 represents a background pixel, and the Maxlable is a preset maximum label value.

Among them, the $\mathrm{Maxlable}=(1+\sqrt{2})\max (m,\mathrm{no}).$

It can be known from the above-mentioned technical solutions that, using the method for fast three-dimensional binary image complete distance transformation provided by the embodiment of the present invention, firstly, in the case of knowing the three-dimensional complete distance transformation of the second target pixel, it can be based on the second target phase pixel in The three-dimensional complete distance transformation value in the three-dimensional binary image determines the range of the first nearest background pixel of the first pixel, and then searches for the nearest background pixel from the first pixel, and searches within the range of the first nearest background pixel instead of Search in the entire binary image, thereby improving the speed of searching for the nearest background pixel; secondly, after searching out the first background pixel closest to the first pixel, then calculate the first background pixel and the first pixel. The complete distance does not calculate the distance between the background pixel and the first pixel after searching out a background pixel, thereby reducing the amount of calculation and improving the calculation speed, and no approximation is performed during the entire search process and calculation process, so The calculated distance has high accuracy.

Description of drawings

In order to more clearly illustrate the technical solutions in the embodiments of the present invention or the prior art, the following will briefly introduce the drawings that need to be used in the description of the embodiments or the prior art. Obviously, the accompanying drawings in the following description are only It is an embodiment of the present invention, and those skilled in the art can also obtain other drawings according to the provided drawings without creative work.

Fig. 1 is a flowchart of the first method for complete distance transformation of a three-dimensional binary image disclosed in an embodiment of the present invention;

Fig. 2 is the clear process diagram of the background pixel proof that the distance O (x, y) is actually the nearest background pixel of the search distance (i, y) in the i-th row in the first function search two-dimensional binary image;

Fig. 3 is a schematic diagram of the second closest background pixel range of the first type;

Fig. 4 is a schematic diagram of the range of the second closest background pixel of the second type;

Fig. 5 is a schematic diagram of the second closest background pixel range of the third type;

Fig. 6 is a proof figure that the background pixel closest to the first pixel in the two-dimensional binary picture is actually the closest background pixel to the projection pixel of the first pixel in the two-dimensional binary picture;

Fig. 7 is a schematic diagram of the first nearest background pixel range;

Fig. 8 is a schematic flowchart of a two-dimensional complete distance transformation method for separately calculating each pixel in each segmented two-dimensional binary image provided by an embodiment of the present invention;

FIG. 9 is a schematic diagram of division of the second closest background pixel;

FIG. 10 is a schematic flowchart of a two-dimensional complete distance transformation method for separately calculating each pixel in each segmented two-dimensional binary image provided by an embodiment of the present invention;

Fig. 11 is a schematic diagram of searching for the nearest background pixel by using the functions LR[x, y] and TD[x, y];

FIG. 12 is a schematic diagram of searching for the nearest background pixel by using the contour search method.

Detailed ways

The following will clearly and completely describe the technical solutions in the embodiments of the present invention with reference to the accompanying drawings in the embodiments of the present invention. Obviously, the described embodiments are only some, not all, embodiments of the present invention. Based on the embodiments of the present invention, all other embodiments obtained by persons of ordinary skill in the art without making creative efforts belong to the protection scope of the present invention.

The x in (x, y) and [x, y] in all embodiments of the present invention represents the row in the two-dimensional binary picture, and does not represent the abscissa, and the x in (x, y) and [x, y] y represents the column in the binary image, not the ordinate. For example, the pixel (x ₀ , y ₀ ) represents the pixel's row x ₀ and column y ₀ in the binary image.

Embodiment one

Please refer to accompanying drawing 1, it is the flow chart of the method for the first kind of three-dimensional binary image complete distance transformation that the embodiment of the present invention discloses, and this method can comprise:

Step S101: dividing the three-dimensional binary image into a corresponding number of two-dimensional binary images along one of the axes;

Specifically, along the x-axis, the three-dimensional binary image can be divided into m two-dimensional binary pictures; along the y-axis, the three-dimensional binary image can be divided into n two-dimensional binary pictures; along the z-axis, the three-dimensional binary The value image is divided into s two-dimensional binary images.

Step S101 is a prior art and will not be described in detail here.

Step S102: Calculate the two-dimensional complete distance transformation of each pixel in each segmented two-dimensional binary image;

The complete distance transformation of each pixel in each two-dimensional binary image can be calculated by using the complete distance transformation method in the prior art.

Preferably, to calculate the distance transformation of a two-dimensional binary picture (assuming that the size of the two-dimensional binary picture is m×n, and m≥1, n≥1), first calculate the complete (1,1) in the first row Distance transformation, and then calculate the complete distance transformation of (1,2), calculate it in turn from left to right to (1,n), and then calculate the complete distance transformation of (2,1) in the second line, from left to right The right is calculated to (2,n) in turn, that is, the complete distance transformation of each pixel in the binary image is calculated in the order from left to right and top to bottom.

Step S102 may include:

S1021: Determine the first function or the second function;

Step S1021 is a preprocessing step and does not need to be executed every time.

The first function is used to determine the position of the background pixel closest to the fourth pixel in row i of the two-dimensional binary image, and the second function is used to determine the position of the background pixel closest to the fourth pixel in column j of the two-dimensional binary image position, where, 1≤i≤m, 1≤j≤n, i and j are both integers.

Assuming that the fourth pixel is located in the xth row and yth column of the two-dimensional binary image, use (x, y) to represent the position of the fourth pixel in the two-dimensional binary image, and use (x, y) as the dividing point, divide the first Row x is divided into left and right, and column y is divided into above and below.

The method for determining the first function specifically includes:

Determining the first sub-function used to calculate the column number of the background pixel closest to the fourth pixel (x, y) on the left side of the xth row of the two-dimensional binary picture, and used to calculate the first sub-function of the two-dimensional binary picture The second subfunction of the column number of the background pixel closest to the fourth pixel (x, y) on the right side of row x. Among them, 1≤x≤m, 1≤y≤n.

The first function is determined according to the first sub-function and the second sub-function.

Determining the second function specifically includes: determining a third sub-function for calculating the row number of the background pixel closest to the fourth pixel (x, y) above the yth column of the two-dimensional binary image, and for calculating the The fourth sub-function of the row number of the background pixel closest to the fourth pixel (x, y) below the y-th column in the two-dimensional binary image.

The second function is determined according to the third sub-function and the fourth sub-function.

Specifically, the first method for determining the first function and the second function is specifically:

The first sub-function is specifically L[x, y]:

$\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; the\; y</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and y=1, then L(x, y)=-Maxlable;

The second sub-function is specifically R[x, y]:

$\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; the\; y</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and y=n, then R[x, y]=Maxlable;

Then the first function determined according to the first sub-function and the second sub-function is specifically SZ[x, y]:

$\mathrm{<span\; class="notranslate">\; SZ</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; <</span>\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}\\ \mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

The third sub-function is specifically T[x, y]:

$\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; x</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and x=1, then T[x, y]=-Maxlable;

The fourth sub-function is specifically D[x, y]:

$\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; x</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and x=m, then D[x, y]=Maxlable;

Then the second function determined according to the third sub-function and the fourth sub-function is specifically ZS[x, y]:

$\mathrm{<span\; class="notranslate">\; ZS</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; <</span>\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; x</span>}\\ \mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; x</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

The second method for determining the first function and the second function is specifically:

The first sub-function is specifically L[x, y]:

$\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; the\; y</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and y=1, then L(x, y)=-Maxlable;

The second sub-function is specifically R[x, y]:

$\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; the\; y</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and y=n, then R[x, y]=Maxlable;

Then the first function determined according to the first sub-function and the second sub-function is specifically LR[x, y]:

$\mathrm{<span\; class="notranslate">\; LR</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; <</span>\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}\\ {<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

The third sub-function is specifically T[x, y]:

$\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; x</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and x=1, then T[x, y]=-Maxlable;

The fourth sub-function is specifically D[x, y]:

$\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; x</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$

If I(x, y)=1 and x=m, then D[x, y]=Maxlable;

Then the second function determined according to the third sub-function and the fourth sub-function is specifically TD[x,y]:

$\mathrm{<span\; class="notranslate">\; TD</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}{<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}& \mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; <</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; x</span>}\\ {<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}& \mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; x</span>}\end{array}\right.<span\; class="notranslate">\; .</span>$

Among them, I(x, y)=1 represents the target pixel, I(x, y)=0 represents the background pixel, and Maxlable is the preset maximum mark value, where, $\mathrm{Maxlable}=(1+\sqrt{2})\max (m,\mathrm{no}).$

S1022: Determine the second closest background pixel range of the fourth pixel according to the complete distance between the third target pixel and the fourth pixel, and the two-dimensional complete distance between the third target pixel and its nearest background pixel.

Wherein, both the fourth pixel and the third target pixel are located in the same two-dimensional binary image;

In order to improve the calculation speed, before determining the second background pixel range, it may also include:

Calculate LR[x,y]; when LR[x,y] is not greater than 1, the first complete distance is LR[x,y], and when LR[x,y] is greater than 1, execute Determine the second closest background pixel range step.

Or, calculate TD[x,y]; when TD[x,y] is not greater than 1, the second complete distance is TD[x,y], and when TD[x,y] is greater than 1, execute Determine the second closest background pixel range step.

If LR(x,y)=0 or TD[x,y]=0, it means that the fourth pixel (x,y) is a background pixel. If LR(x, y)=1, it means that there is a background pixel in the horizontal 2 neighborhood of the fourth pixel (x, y), that is, the second complete distance is 1; if TD[x, y]=1, it means that the fourth pixel (x, y) has a background pixel There are background pixels in the vertical 2-neighborhood of pixel (x,y), i.e. the second complete distance is 1.

Before determining the second background pixel range, it may also include:

Calculate (y-SZ[x,y]) ² ; when (y-SZ[x,y]) ² is not greater than 1, the second complete distance is (y-SZ[x,y]) ² , in When (y-SZ[x,y]) ² is greater than 1, perform the step of determining the second nearest background pixel range.

Or, calculate (x-ZS[x,y]) ² ; if (x-ZS[x,y]) ² is not greater than 1, the second complete distance is (x-ZS[x,y]) ² , when (x-ZS[x,y]) ² is greater than 1, execute the step of determining the second nearest background pixel range.

Use the first function to search for the background pixel closest to the fourth pixel in each row of the second closest background pixel range, or use the second function to search for the closest background pixel to the fourth pixel in each column of the second closest background pixel range background pixels;

Determine the second background pixel closest to the fourth pixel from the searched background pixels closest to the fourth pixel, calculate the second complete distance between the second background pixel and the fourth pixel, and use the second complete distance as the first The 2D full distance transformed value of the quadpixel in its 2D binary image.

The fourth pixel may be any pixel located in the two-dimensional binary image, and any pixel is called the fourth pixel only for the convenience of description.

When the fourth pixel is a background pixel, the background pixel closest to the fourth pixel is itself.

Assuming that the position of the fourth pixel in the two-dimensional binary image is (x, y), since the first sub-function is used to calculate the distance from the left side of row x of the two-dimensional binary image to any pixel (x, y) closest The number of columns of the background pixels of the two-dimensional binary image, the second sub-function is used to calculate the number of columns of the background pixel that is closest to any pixel (x, y) on the right side of row x of the two-dimensional binary image, so according to the first sub-function and The SZ[x, y] obtained by the second sub-function is also used to determine the column number of the background pixel closest to any pixel (x, y) on the xth row.

Use O(x,y) to represent the above fourth pixel, use the first function to search for the background pixel closest to O(x,y) in the i-th row (i≠x), in fact, use the first function to search for the i-th row The background pixel closest to (i, y), so when step S102 is executed, the function SZ[i, y] that replaces x in the first function SZ[x, y] with i is actually searched, and the function SZ[ i, y] actually searches for the background pixel closest to (i, y) in line i, since the background pixel closest to (i, y) in line i is the closest to O(x, y) Background pixels, so the function SZ[i,y] is essentially used to search for the background pixel closest to O(x,y) in each row.

As shown in Figure 2, the proof process is as follows: Suppose L ₁ , L ₂ , R ₁ , R ₂ are the four background pixels on the i-th row, and their distances from O(x, y) are |OL ₁ |, |OL ₂ |, |OR ₁ |, |OR ₂ |, it can be seen from geometric knowledge that the background pixel closest to O(x,y) is obviously the background pixel closest to (i,y), and the distance from O( x,y) The position of the nearest background pixel is (i,SZ[i,y]).

Similarly, using the second function to search for the background pixel closest to O(x, y) in the jth column (j≠y) is actually using the second function to search for the background pixel closest to (x, j) in the jth column , so when step S102 is executed, the function ZS[x,j] that replaces y in the second function ZS[x,y] with j is actually used for searching, and the closest distance to O(x,y) in the jth column The position of the background pixel is (ZS[x,j],j). Wherein, 1≤i≤m, 1≤j≤n, i and j are both positive integers.

Determining the second nearest background pixel range specifically includes:

First, calculate the two-dimensional complete distance r ₃ between the third target pixel O ₃ and the nearest background pixel B ₀ ;

Second, determine the complete distance r ₄ between the fourth pixel O ₄ and the third target pixel O ₃ ;

The calculated complete distance r ₃ can be calculated by methods in the prior art. There may be multiple background pixels closest to the third target pixel O ₃ , and only one of the background pixels B ₀ can be selected according to the actual situation.

Preferably, the fourth pixel O ₄ is located within the four-neighborhood of the third target pixel O ₃ , so that the complete distance r ₄ =1 between the fourth pixel O ₄ and the third target pixel O ₃ . Of course, the position of the fourth pixel _O4 is not limited to the four areas of the third target pixel _O3 , and the specific position of the fourth pixel _O4 does not affect the implementation of the embodiment of the present invention, so the specific position of the fourth pixel _O4 is not discussed. The location is specifically limited.

Finally, the second nearest background pixel range of the fourth pixel _O4 is determined according to the complete distance _r3 and the complete distance _r4 .

Referring to Fig. 3, it is a schematic diagram of the range of the second closest background pixel of the first type. The circle O ₄ in the figure is a circle O ₄ with the fourth pixel as the center and r ₃ +r ₄ as the radius. Due to the discrete nature of digital images, It is very time-consuming to determine the exact circular area bounded by the arc, so the second closest background pixel range is the area enclosed by the circumscribed square of the circle O ₄ .

Referring to FIG. 4 , it is a schematic diagram of the second closest background pixel range of the second type. The second closest background pixel range is the inscribed square and circle of the circle _O5 with the third target pixel _O3 as the center and _r3 as the radius The area enclosed by the circumscribed square of O ₄ ;

Referring to FIG. 5 , it is a schematic diagram of the second closest background pixel range of the third type. The second closest background pixel range is an annular area surrounded by the circumscribed square of circle _O6 and the inscribed square of circle _O4 .

Circle O ₆ is a circle with the fourth pixel as the center and |r ₃ -r ₄ | as the radius.

If the fourth pixel is located at the edge position of the two-dimensional binary image, when determining the second nearest background pixel range, part of the initially determined second closest background pixel range may exceed the two-dimensional binary image, and at this time the determined The intersection of the initial second closest background pixel range and the two-dimensional binary image is used as the second closest background pixel range.

Step S103: According to the complete distance between the second target pixel and the first pixel, and the three-dimensional complete distance transformation value of the second target pixel in the three-dimensional binary image, determine the first nearest background pixel range of the first pixel;

The second target pixel is any target pixel in the three-dimensional binary image.

Determining the first closest background pixel range may include:

Determine the three-dimensional complete distance transformation value r ₁ of the second target pixel in the three-dimensional binary image;

determining the complete distance r ₂ between the first pixel O ₁ and the second target pixel O ₂ ;

The first closest background pixel range is the area surrounded by the circumscribed cube of _the ball O ₁ with the first pixel O 1 as the center and r ₁ +r ₂ as the radius;

Or, the first closest background pixel range is the area surrounded by the inscribed cube of the ball _O2 and the circumscribed cube of the ball _O1 with _the second target pixel O2 as the center of the sphere and _r1 as the radius;

Or, the range of the first closest background pixel is defined by the inscribed cube of the ball _O3 with the first pixel _O1 as the center and the radius | _r1 - _r2 | and the circumscribed cube of the ball _O1 enclosed area.

If the first pixel is located at the edge of the three-dimensional binary image, when determining the first nearest background pixel range, part of the initially determined first closest background pixel range may exceed the three-dimensional binary image. An intersection of the closest background pixel range and the three-dimensional binary image is used as the first closest background pixel range.

Step S104: Determine the two-dimensional complete distance transformation values of the projected pixels of the first pixel in each two-dimensional binary picture in the range of the first closest background pixel;

Please refer to Figure 6, assuming that pixel O(x,y,z) is a target pixel on the two-dimensional binary image z, pixel O ₀ (x,y,z ₀ ) is O(x,y,z) in two The projection pixel on the two-dimensional binary image z ₀ is set to the background pixel B ₀ closest to the pixel O ₀ (x _, y,z ₀ ) and the pixel O ₀ (x,y,z ₀ ) on the two-dimensional binary image z 0 ) is |O ₀ B ₀ | ² , then the background pixel closest to pixel O(x,y,z) on the two-dimensional binary image z ₀ is obviously also B ₀ , and $\left|{\mathrm{OB}}_0\right|=\sqrt{{\left|o_0{B}_0\right|}^2+{\left|o{o}_0\right|}^2}.$

Assuming that the 3D binary image is divided into s two-dimensional binary images along the z-axis, the range of the first closest background pixel is defined by the circumscribed cube of the ball O ₁ with the first pixel O 1 as the center and r ₁ +r ₂ as _the radius For the enclosed area, please refer to Figure 7, which is a schematic diagram of the range of the first nearest background pixel, r=r ₁ +r ₂ in Figure 7, B is a pixel on the ball O ₁ , O ₁ is represented by O in the figure, It can be seen from Figure 7 that the range of the first closest background pixel can be regarded as being formed by superimposing 2(r ₁ +r ₂ )+1 pictures along the z-axis, assuming the two-dimensional binary picture where the first pixel O ₁ is located is z ₀ , preferably, first determine the two-dimensional complete distance transformation value of the first pixel in the two-dimensional binary image z ₀ (at this time, the projected pixel of the first pixel on the two-dimensional binary image z ₀ is itself), Secondly, determine the two-dimensional complete distance transformation value of the first pixel projected in the two-dimensional binary image z ₀ ±1, and then determine the two-dimensional complete distance transformation value of the first pixel projected in the two-dimensional binary image z ₀ ±2 The distance transformation value until the two-dimensional binary image z ₀ ±(r ₁ +r ₂ ). That is, take the two-dimensional binary picture z ₀ as the symmetric center and determine the two-dimensional complete distance transformation values of the projected pixels of the first pixel in each two-dimensional binary picture sequentially along the z-axis direction from the inside to the outside. The square of the two-dimensional complete distance change value of the projected pixel of the first pixel in the two-dimensional binary image z ₀ +r _x determined on the value picture z ₀ +r _x is DR _x ² , r _x ∈ [-(r ₁ +r ₂ ),r ₁ +r ₂ ], since in the two-dimensional binary image z ₀ ±(r _x +1), the minimum value of the square of the distance between the background pixel closest to the first pixel and the first pixel is (r _x +1) ² , so when DR _x ² +r _x ² ≤(r _x +1) ² , stop determining the first pixel in the two-dimensional binary image z ₀ ±(r _x +2) until two-dimensional two value picture z ₀ ±(r ₁ +r ₂ ) the two-dimensional complete distance transformation value of the projection pixel, if DR _x ² +r _x ² >(r _x +1) ² , continue to determine the first pixel in the two The two-dimensional complete distance transformed values of the projected pixels in the value image z ₀ ±(r _x +2) until the two-dimensional binary image z ₀ ±(r ₁ +r ₂ ).

The range of the first closest background pixel can be regarded as the superposition of 2(r ₁ +r ₂ )+1 pictures along the z-axis, and it is very likely that the projected pixel closest to the first pixel in a two-dimensional binary picture The background pixel is not within the range of the first closest background pixel, but this does not affect the implementation of the embodiment of the present invention, so no distinction is made here.

Step S105: Determine the first background closest to the first pixel in the three-dimensional binary image according to the determined two-dimensional complete distance transformation values of the first pixels in the first background pixels of the projected pixels of each two-dimensional binary image pixels, calculating a first complete distance between the first background pixel and the first pixel, and using the first complete distance as a three-dimensional complete distance transformation value of the first pixel.

Suppose the position of the first pixel in the three-dimensional binary image is (x, y, z), since the two-dimensional complete distance of each pixel in each two-dimensional binary image has been calculated, assuming that DR _x represents the distance from the two-dimensional The square value of the background pixel closest to the first pixel in the two-dimensional binary image z _x in the binary image z _x , where the superscript x indicates that the projected pixel of the first pixel is on the two-dimensional binary image z _x , Then step S105 specifically includes:

Using the formula min{DR _x ² +|z _x -z| ² }, where z _x ∈ the value range of the z-axis in the range of the first nearest background pixel, determine the first background pixel closest to the first pixel, and then use the formula Get the first complete distance.

Preferably, when calculating the three-dimensional complete distance transformation of each pixel in each two-dimensional binary image, it is assumed that the three-dimensional binary image is divided into s two-dimensional binary images along the z-axis, and the range of these s two-dimensional binary images in the z-axis is [1, s], then first calculate the three-dimensional complete distance transform of each pixel in each two-dimensional binary picture z=1 from the two-dimensional binary picture z=1, secondly, calculate the three-dimensional complete distance transformation of each pixel in the two-dimensional binary picture z=2 Distance transformation, and then calculate the three-dimensional complete distance transformation of each pixel in the two-dimensional binary image z=3, and so on until the three-dimensional complete distance transformation of each pixel in the two-dimensional binary image z=s is calculated, that is, the two-dimensional binary image z=s The minimum value of z in the picture is calculated to the maximum value of z.

Using the method for fast three-dimensional binary image complete distance transformation provided by the embodiment of the present invention, firstly, under the condition that the three-dimensional complete distance transformation of the second target pixel is known, the three-dimensional distance transformation of the second target phase pixel in the three-dimensional binary image can be Complete distance transformation value, determine the first closest background pixel range of the first pixel, and then search for the closest background pixel to the first pixel, it is searched within the first closest background pixel range, not in the entire binary image Search, thereby improving the speed of searching for the nearest background pixel; secondly, after searching out the first background pixel closest to the first pixel, then calculating the first complete distance between the first background pixel and the first pixel is not to search for the first background pixel After one background pixel, the distance between the background pixel and the first pixel is calculated, thereby reducing the amount of calculation and improving the calculation speed, and no approximation is performed during the entire search process and calculation process, so the calculated distance is accurate high.

Embodiment two

Please refer to FIG. 8 , which is a schematic flowchart of a two-dimensional complete distance transformation method for separately calculating each pixel in each segmented two-dimensional binary image provided by the embodiment of the present invention, that is, the implementation method of step S102 in the first embodiment, The method can include:

Step S801: According to the complete distance between the third target pixel and the fourth pixel, and the distance between the third target pixel and its nearest background pixel, determine the second closest background pixel range of the fourth pixel, and set the second closest background pixel range According to preset rules, it is divided into a first sub-nearest background pixel range set and a second sub-nearest background pixel range set, the first sub-nearest background pixel range set includes at least one first sub-nearest background pixel range, and the at least one sub-nearest background pixel range The number of rows in the pixel range is not greater than the number of columns, the second sub-nearest background pixel range set includes at least one second sub-nearest background pixel range, and the number of rows in the at least one second sub-nearest background pixel range is greater than the number of columns;

Before step S801, it also includes: calculating LR[x, y] and TD[x, y]; in LR[x, y] or TD[x, y]

When not greater than 1, the second complete distance is min{LR[x,y],TD[x,y]}, in LR[x,y]

And when both TD[x, y] are greater than 1, go to step S802.

If LR(x,y)=0 or TD[x,y]=0, it means that the fourth pixel (x,y) is a background pixel. If LR(x,y)=1, it means that there is a background pixel in the horizontal 2 neighborhood of the fourth pixel (x,y), that is, the second complete distance is 1, and if TD[x,y]=1, it means that the fourth There are background pixels in the vertical 2-neighborhood of pixel (x,y), i.e. the second complete distance is 1.

The second nearest background pixel range as shown in Figure 4 or Figure 5, extends the line segment of a pair of sides inscribed in the square to intersect with the boundary of the circumscribed square, here the second nearest The background pixel range is divided as an example, as shown in Figure 9, which is a schematic diagram of the division of the second nearest background pixel, the dotted line is the extension of a pair of sides of a square inscribed in the circle _O3 , and the pair of sides divides the second nearest background pixel The range is divided into four sub-closest background pixel ranges, namely upper, lower, left, and right areas. Obviously, the number of rows in the upper and lower areas is less than the number of columns, and in the left and right areas The number of columns is less than the number of rows. Of course, the extension line of the other opposite side of the square may also be inscribed by the circle O ₃ to divide the range of the nearest background pixel into four areas: upper, lower, left, and right.

The default rule is to extend the line segment of a pair of sides of the inscribed square of the circle _O3 to intersect with the boundary of the circumscribed square. The pair of sides divides the second closest background pixel range into four sub-closest background pixel ranges: upper, lower, left, and right.

The first sub-nearest background pixel range set includes: an upper area and a lower area; the second sub-nearest background pixel range set includes: a left area and a right area.

Step S802: Use the first function to search for the background pixel closest to the fourth pixel in each row of the first sub-nearest background pixel range set, and use the second function to search for the background pixel closest to the fourth pixel in each column of the second sub-nearest background pixel range set. the four-pixel nearest background pixel;

When the number of rows in the first sub-nearest background pixel range set is not greater than the number of columns, use the first function to search, and the search time is less; when the number of rows in the second sub-nearest background pixel range set is greater than the number of columns, use the second function The second function is used to search, and the search time used is less.

Step S803: Determine the second background pixel closest to the fourth pixel from the searched background pixels closest to the fourth pixel, calculate the second complete distance between the second background pixel and the fourth pixel, and calculate the second full distance as the 2D full distance transformed value of the fourth pixel.

According to the embodiment of the present invention, since the second nearest background pixel range is divided according to predetermined rules, since the number of rows of the first sub-nearest background pixel range set is not greater than the number of columns, the first function is used in the first sub-nearest background pixel range Search for the background pixel closest to the fourth pixel in each row of the set, because the number of rows in the second sub-nearest background pixel range set is greater than the number of columns, use the second function to search for the distance in each column of the second sub-nearest background pixel range set The nearest background pixel of the fourth pixel accelerates the search speed, thereby improving the calculation speed.

Embodiment Three

Please refer to FIG. 10 , which is a schematic flowchart of a two-dimensional complete distance transformation method for separately calculating each pixel in each segmented two-dimensional binary image provided by the embodiment of the present invention, that is, the implementation method of step S102 in the first embodiment. Methods can include:

Step S1001: According to the complete distance between the third target pixel and the fourth pixel, and the two-dimensional complete distance between the third target pixel and its nearest background pixel, determine the second closest background pixel range of the fourth pixel;

Step S1002: using the first function and the second function to search for the background pixel closest to the fourth pixel in each layer of the second closest background pixel range;

Each layer of the second closest background pixel range refers to a boundary of a square centered on the fourth pixel or the third target pixel.

Specifically, use the first function and the second function to start searching from the _rxth layer until the _rx +kth layer searches for the background pixel closest to the fourth pixel, and the _rxth layer is the fourth pixel within the range of the nearest background pixel. The nearest layer of pixels, k=0,1,2...q-1, there are a total of q layers within the range of the second closest background pixel;

From the background pixels closest to the fourth pixel searched in the _rx +kth layer, determine a background pixel in the _rx +kth layer closest to the fourth pixel, and record all the background pixels in the _rx +kth layer The complete distance square value between a background pixel closest to the fourth pixel and the fourth pixel

Compare with (r _x +k+1) ² +(r _x +k) ² , when , stop searching when When , continue to search the next layer until the r _x + q-1th layer is finished.

Taking the second closest background pixel range as shown in Figure 5 as an example, step S1002 is described, each layer of the second closest background pixel is the boundary of a square centered on O ₄ , and half of the square side length is For the number of layers of this layer, the method of searching within the nearest background pixel range by using the idea of perimeter search is as follows: Layer (half the side length of the square inscribed in the circle) starts, and scans layer by layer until r ₁ + r ₂ layers (half the side length of the square circumscribed by the circle) end, assuming that the r _xth layer is scanned, where Referring to Fig. 11, it is a schematic diagram of using the functions LR[x, y] and TD[x, y] to search for the nearest background pixel. For the pixels on the r _x layer, they are divided into R ₁ along the x and y directions respectively (in the figure The area surrounded by dotted lines), R ₂ (the area enclosed by dotted lines in the figure), R ₃ (the area surrounded by the left side of the figure) and R ₄ (the area surrounded by the right side of the figure) It is easy to know that the elements in the four groups are (xr _x ) row, (x+r _x ) row, (yr _x ) column, and part of the pixel on (y+r _x ) column. The coordinates of the background pixel (denoted as B _x1 ) closest to the target pixel O in the (xr _x ) row are or like It indicates that B _x1 ∈ R ₁ , that is, there are background pixels in R _1. Similarly, if LR[x+r _x ,y]<r _x ² ,TD[x,yr _x ]<r _x ² TD[x _x ,y +r]<r _x ² , it indicates that there are background pixels in R ₂ , R ₃ , and R ₄ respectively. Let tD=min{LR[xr _x ,y],LR[x+r _x ,y],TD[x ,yr _x ],TD[x _x ,y+r]}, record the square of the distance between the background pixel closest to the fourth pixel on layer r _x and the fourth pixel as DT _x ² =tD+r _x ² , because in The minimum value of tD in the r _x +1th layer is (r _x +1) ² , so when tD≤(r _x +1) ² , stop searching, or judge when to stop searching.

Referring to Figure 12, it is a schematic diagram of searching for the nearest background pixel by using the contour search method, assuming that the fourth pixel that needs to calculate the distance transformation is the target pixel A, the number 1 in the figure represents the first layer, the number 2 represents the second layer, and the number 3 represents The third layer and the number 4 represent the fourth layer. Only four layers are drawn in the figure, but it is not necessarily four layers in practical applications. Assume that when r _x =2, a background pixel B ₄ is found, obviously the background pixel The distance between B ₄ and target pixel A is but The square value of 18 is greater than (3+1) ² , so it is necessary to continue searching the third layer. When r _x =3, two background pixels B ₁ and B ₃ are found. Due to the distance between the background pixel B ₁ and the target pixel A for Obviously, 13 is less than (4+1) ² , so the search for the next layer is stopped. At this time, the background pixels closest to the first pixel searched by the first function and the second function in the range of the nearest background pixels are B ₁ , _B3 and _B4 .

The above is an example of _the nearest background pixel range shown in Fig. 4, to explain step S1002, if the second closest background pixel is as shown in Fig. The boundary of the square in the center, then the method of searching within the nearest background pixel range using the idea of perimeter search is as follows: start from the first layer (that is, start from the 8th area of the center of the circle), and scan layer by layer until r ₁ + r ₂ layers (half the side length of the square circumscribed by the circle); if the second closest background pixel is as shown in Figure 3, and each layer of the second closest background pixel range refers to a square with O ₃ as the center, then use the perimeter The idea of searching The method of searching within the nearest background pixel range is as follows: starting from the first layer (that is, starting from the 8 field of the center of the circle), when scanning to the _r1th layer (the _r1th layer is the circumscribed circle _O3 half of the side length of the square), the left side of the second closest background pixel range has been searched, and each layer of the second closest background pixel range is bounded by a dotted line box centered on O ₃ as shown in Figure 3 , search from layer r ₁ +1 until layer r ₁ +r ₂ .

Step S1003: Determine the second background pixel closest to the fourth pixel from the searched background pixels closest to the fourth pixel, calculate the second complete distance between the second background pixel and the fourth pixel, and calculate the second full distance as the 2D full distance transformed value of the fourth pixel.

The complete distance refers to the shortest distance between the second background pixel and the fourth pixel, that is, the straight-line distance, and the distance transformation value refers to the distance value between the background pixel closest to the target pixel and the target pixel.

Specifically, step S1003 includes: according to the recorded complete distance square value between the background pixel closest to the fourth pixel and the fourth pixel, determine the distance between the second background pixel closest to the fourth pixel and the fourth pixel The square value of the second complete distance; the square root of the square value of the second complete distance is used to obtain the second complete distance.

The square of the distance between the background pixel closest to the fourth pixel and the fourth pixel on the r _x +k layer is Assuming that only the background pixel closest to the fourth pixel is found on the third layer, the fourth layer and the fifth layer, step S1003 specifically includes:

use the formula Determine the second background pixel closest to the fourth pixel, and then use the formula: Calculate the second complete distance.

The method provided by the embodiment of the present invention not only has the beneficial effect of the first embodiment, but also uses the first function and the second function to search in each layer of the nearest background pixel, so the search speed is faster.

In order for those skilled in the art to further understand the advantages of the embodiments of the present invention, the inventor uses the above method to calculate the complete distance transformation of each pixel in the 3D binary image, and uses the existing technology to perform complete distance transformation on the same 3D binary image, and Compare the results of the two operations.

The prior art refers to the method described in Euclidean distance transform of digital images in arbitrary dimensions. PCM 2006, LNCS, 2006, 4261:72-79.

As shown in Table 1, it is a comparison table of the running time (unit: second) for calculating the three-dimensional complete distance transformation in the prior art and the time for calculating the three-dimensional complete distance transformation using this method.

Table 1

It can be seen from Table 1 that for distance transformation of images of different sizes, the time required for the existing technology to perform three-dimensional complete distance transformation is longer than that of the method provided by the embodiment of the present invention. In order to express the embodiment of the present invention more intuitively Based on the idea that the provided method for performing three-dimensional complete distance transformation has high efficiency, Table 1 also shows the running time ratio of "the prior art/the method provided by the embodiment of the present invention".

Each embodiment in this specification is described in a progressive manner, each embodiment focuses on the difference from other embodiments, and the same and similar parts of each embodiment can be referred to each other. As for the device disclosed in the embodiment, since it corresponds to the method disclosed in the embodiment, the description is relatively simple, and for the related information, please refer to the description of the method part.

The steps of the methods or algorithms described in connection with the embodiments disclosed herein may be directly implemented by hardware, software modules executed by a processor, or a combination of both. Software modules can be placed in random access memory (RAM), internal memory, read-only memory (ROM), electrically programmable ROM, electrically erasable programmable ROM, registers, hard disk, removable disk, CD-ROM, or any other Any other known storage medium.

The above description of the disclosed embodiments is provided to enable any person skilled in the art to make or use the invention. Various modifications to these embodiments will be readily apparent to those skilled in the art, and the general principles defined herein may be implemented in other embodiments without departing from the spirit or scope of the invention. Therefore, the present invention will not be limited to the embodiments shown herein, but is to be accorded the widest scope consistent with the principles and novel features disclosed herein.

Claims (8)

1. A method for complete distance transformation of a three-dimensional binary image, the size of the three-dimensional binary image is m × n × s, characterized in that, comprising: dividing the three-dimensional binary image into s two-dimensional binary images along the z axis; Calculating the two-dimensional complete distance transform of each pixel in each segmented two-dimensional binary image; According to the complete distance between the second target pixel and the first pixel for which three-dimensional complete distance transformation needs to be calculated, and the three-dimensional complete distance transformation value of the second target pixel in the three-dimensional binary image, determine the first pixel of the first pixel. a nearest background pixel range; determining the two-dimensional complete distance transform value of the projected pixel of the first pixel in each two-dimensional binary image within the range of the first nearest background pixel; According to the determined two-dimensional complete distance transformation value of the projected pixel of each two-dimensional binary picture in the first closest background pixel range of the first pixel, determine the distance from the first pixel in the three-dimensional binary image The nearest first background pixel, calculating the first complete distance between the first background pixel and the first pixel, and using the first complete distance as the three-dimensional complete distance transformation value of the first pixel; The 3D complete distance transformation value of the second target pixel in the 3D binary image is r ₁ , according to the complete distance between the second target pixel and the first pixel, and the 3D distance of the second target pixel in the 3D The three-dimensional complete distance transformation value in the binary image, determining the first nearest background pixel range of the first pixel specifically includes: determining a full distance r ₂ of the first pixel to the second target pixel; The first closest background pixel range is the area surrounded by a circumscribed cube of a ball O ₁ with the first pixel as the center and r ₁ +r ₂ as the radius; or, the first closest background pixel The range is the area enclosed by the inscribed cube of the ball _O2 and the circumscribed cube of the ball _{O1 with the second target pixel as the center and r1 as} the radius; or, the first closest background The pixel range is the area surrounded by the inscribed cube of the ball O ₃ and the circumscribed cube of the ball O ₁ with the first pixel as the center and |r ₁ -r ₂ | as the radius; When the first closest background pixel range is the area surrounded by a circumscribed cube of a ball O ₁ with the first pixel as the center and r ₁ +r ₂ as the radius, the first pixel is located The two-dimensional binary picture is z ₀ , and the determination of the two-dimensional complete distance transformation value of the first pixel in each two-dimensional binary picture in the range of the first nearest background pixel specifically includes: Taking the two-dimensional binary picture _z0 as the center of symmetry and sequentially determining the two-dimensional complete distance transformation value of the projected pixel of the first pixel in the two-dimensional binary picture _z0 + _rx from the inside to the outside along the z-axis direction The square of DR _x ² , and record the square of the determined two-dimensional complete distance transformation value, r _x =0,±1,±2,...,±(r ₁ +r ₂ ); Judging whether DR _x ² satisfies DR _x ² +r _x ² ≤(r _x +1) ² , if yes, stop determining the first pixel in the two-dimensional binary image z ₀ ±(r _x +1) to two-dimensional The two-dimensional complete distance transformation value of the projected pixel in the binary image z ₀ ±(r ₁ +r ₂ ), if not, continue to determine that the first pixel is respectively in the two-dimensional binary image z ₀ ±(r _x +1) Projecting the two-dimensional complete distance transformation value of the pixel until the two-dimensional complete distance transformation value of the projected pixel of the first pixel in the two-dimensional binary image z ₀ ±(r ₁ +r ₂ ) is determined. 2. The method according to claim 1, wherein, according to the two-dimensional complete distance transformation values of the projected pixels of each two-dimensional binary picture in the first closest background pixel range of the first pixel, determine Finding the first background pixel closest to the first pixel in the three-dimensional binary image specifically includes: use the formula Wherein z _x ∈ the value range of z in the first nearest background pixel range, determine the first background pixel closest to the first pixel distance; use the formula Calculate the first complete distance. 3. The method according to claim 1, wherein calculating the two-dimensional complete distance transformation of each two-dimensional binary picture specifically comprises: A preprocessing step, the preprocessing step is to determine the first function or determine the second function, the first function is used to determine the position of the background pixel closest to the fourth pixel in the i-th row of the two-dimensional binary image, the The second function is used to determine the position of the background pixel closest to the fourth pixel in the jth column of the two-dimensional binary image, wherein, 1≤i≤m, 1≤j≤n, i and j are both integers; According to the complete distance between the third target pixel and the fourth pixel for which the two-dimensional complete distance needs to be calculated, and the two-dimensional complete distance of the third target pixel in the two-dimensional binary image, determine the first position of the fourth pixel 2. The closest background pixel range, the fourth pixel and the third target pixel are located in the same two-dimensional binary image; Use the first function to search for the background pixel closest to the fourth pixel in each row of the second closest background pixel range, or use the second function to search for the background pixel in the second closest background pixel range Search for the nearest background pixel to the fourth pixel in each column; Determine the second background pixel closest to the fourth pixel from the searched background pixels closest to the fourth pixel, and calculate the second two-dimensional completeness between the second background pixel and the fourth pixel distance, and use the second two-dimensional complete distance as the two-dimensional complete distance transformation value of the fourth pixel in the two-dimensional binary picture where it is located. 4. The method according to claim 1, wherein calculating the two-dimensional complete distance transformation of each two-dimensional binary picture specifically comprises: A preprocessing step, the preprocessing step is to determine the first function or determine the second function, the first function is used to determine the position of the background pixel closest to the fourth pixel in the i-th row of the two-dimensional binary image, the The second function is used to determine the position of the background pixel closest to the fourth pixel in the jth column of the two-dimensional binary image, wherein, 1≤i≤m, 1≤j≤n, i and j are both integers; According to the complete distance between the third target pixel and the fourth pixel, and the distance between the third target pixel and its nearest background pixel, determine the second closest background pixel range of the fourth pixel, and divide the second closest background pixel range according to The preset rule is divided into a first sub-nearest background pixel range set and a second sub-nearest background pixel range set, the first sub-nearest background pixel range set includes at least one first sub-nearest background pixel range, and the at least one first sub-nearest background pixel range The number of rows in the background pixel range is not greater than the number of columns, the second sub-nearest background pixel range set includes at least one second sub-nearest background pixel range, and the number of rows in the at least one second sub-nearest background pixel range is greater than the number of columns; Use the first function to search for the background pixel closest to the fourth pixel in each row of the first sub-nearest background pixel range set, and use the second function to search for the closest background pixel to the fourth pixel in each column of the second sub-nearest background pixel range set background pixels; Determine the second background pixel closest to the fourth pixel from the searched background pixels closest to the fourth pixel, calculate the second complete distance between the second background pixel and the fourth pixel, and use this second complete distance as The 2D full distance transformed value of the fourth pixel. 5. The method according to any one of claims 3 or 4, wherein said determining the second nearest background pixel range of said fourth pixel specifically comprises: Calculating the two-dimensional complete distance r ₃ of the third target pixel in the two-dimensional binary image; determining a complete distance r ₄ of the fourth pixel from the third target pixel; The second closest background pixel range is the area surrounded by the circumscribed square of the circle O ₄ with the fourth pixel as the center and r ₃ +r ₄ as the radius; Or, the second closest background pixel range is an annular area surrounded by the inscribed square of the circle _O5 and the circumscribed square of the circle _O4 with the third target pixel as the center and _r3 as the radius ; Or, the range of the second closest background pixel is defined by the inscribed square of the circle O ₆ with the fourth pixel as the center and the radius |r ₃ -r ₄ | and the circumscribed square of the circle O ₄ enclosed circular area. 6. according to any described method of claim 3 or 4, it is characterized in that, described 4th pixel is positioned at the xth line of two-dimensional binary picture, yth column, represents described 4th pixel with (x, y) In the position in the two-dimensional binary picture, with (x, y) as the dividing point, the xth row is divided into left and right sides, and the yth column is divided into top and bottom; The determining the first function specifically includes: Determining the first sub-function used to calculate the column number of the background pixel closest to the fourth pixel (x, y) on the left side of the xth row of the two-dimensional binary image, and used to calculate the two-dimensional binary image The second sub-function of the column number of the background pixel closest to the fourth pixel (x, y) on the right side of the xth row of the picture, wherein, 1≤x≤m, 1≤y≤n; determining the first function according to the first sub-function and the second sub-function; The determining the second function specifically includes: Determine the third sub-function used to calculate the row number of the background pixel closest to the fourth pixel above the yth column of the two-dimensional binary image, and to calculate the distance below the yth column of the two-dimensional binary image The fourth sub-function of the row number of the nearest background pixel of the fourth pixel; The second function is determined according to the third sub-function and the fourth sub-function. 7. The method according to claim 6, wherein the first sub-function is specifically L[x, y]: $\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; the\; y</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ If I(x, y)=1 and y=1, then L[x, y]=-Maxlable; The second sub-function is specifically R[x, y]: $\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; the\; y</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ If I(x, y)=1 and y=n, then R[x, y]=Maxlable; Then the first function determined according to the first sub-function and the second sub-function is specifically SZ[x, y]: $\mathrm{<span\; class="notranslate">\; SZ</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; <</span>\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}\\ \mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; L</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; R</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ The third sub-function is specifically T[x, y]: $\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; x</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ If I(x, y)=1 and x=1, then T[x, y]=-Maxlable; The fourth sub-function is specifically D[x, y]: $\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; x</span>}& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 0</span>}\\ \mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; +</span>\mathrm{<span\; class="notranslate">\; 1</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; I</span>}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>\mathrm{<span\; class="notranslate">\; 1</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ If I(x, y)=1 and x=m, then D[x, y]=Maxlable; Then the second function determined according to the third sub-function and the fourth sub-function is specifically ZS[x, y]: $\mathrm{<span\; class="notranslate">\; ZS</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; =</span>\left\{\begin{array}{cc}\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; <</span>\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; x</span>}\\ \mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span>& \mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; D.</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; \&Greater\; Equal;</span>\mathrm{<span\; class="notranslate">\; T</span>}<span\; class="notranslate">\; [</span>\mathrm{<span\; class="notranslate">\; x</span>}<span\; class="notranslate">\; ,</span>\mathrm{<span\; class="notranslate">\; the\; y</span>}<span\; class="notranslate">\; ]</span><span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; x</span>}\end{array}\right.<span\; class="notranslate">\; ;</span>$ Wherein, I(x, y)=1 represents a target pixel, I(x, y)=0 represents a background pixel, and the Maxlable is a preset maximum label value. 8. The method according to claim 7, wherein the