CN101930666A - Testing method of one-way road section traffic volume composition - Google Patents

Abstract

The invention discloses a testing method of one-way road section traffic volume composition comprising the following steps of initializing traffic demand; initializing city road network; initializing impedance, traffic volume and traffic volume composition of each one-way road section; initializing cycle index i which is equal to 1; judging whether i is not more than n; if the i is not more than n, executing an impedance matrix generation algorithm, calculating the current impedance matrix by using the impedance matrix generation algorithm and preparing for searching the shortest path in traffic allocation by using any one node as a source point and a preposing node relative to the other node, and if the i is more than 1 ending to obtain the traffic volume composition on each road section; allocating the traffic demand ODi matrix to a previous directed network Gi-1(V, E) by using a traffic allocation algorithm to form a current directed network Gi (V, E); recording the traffic volume composition on each one-way road section of the directed network Gi (V, E), wherein the traffic volume composition comprises a starting point, an ending point and volume; supposing i=i+1, continuously judging whether the i is not more than n, and circularly executing until all OD matrixes are allocated; accumulating all traffic volumes on the one-way road section according to the starting point and the ending point to form final one-way road section traffic volume composition which is represented in a form of table.

Description

The assay method that one-way road section traffic volume constitutes

Technical field

The present invention relates to the technical field of traffic forecast and evaluation method, relate in particular to the assay method that a kind of road section traffic volume constitutes.

Background technology

This method is the quadravalence section to traditional traffic forecast " quadravalence section " method---the improvement that traffic distributes.Traffic forecast " quadravalence section " method distributes four parts to constitute by plot generation and traffic attraction, OD distribution, mode of transportation division and traffic.This method is by the improvement to the conventional traffic assigning process, and the volume of traffic of having analyzed each highway section constitutes, and each volume of traffic by which plot produces, attracts and amount, represents with the form of form at last.

Summary of the invention

1, technical matters:

The invention provides the assay method that a kind of one-way road section traffic volume constitutes, the advantage that can provide each volume of traffic in the unidirectional highway section that origin and destination and amount thereof take place is provided.

2, technical scheme:

The assay method that a kind of one-way road section traffic volume constitutes:

Step 1: the initialization transport need on average is split as ∏ five equilibrium: OD with the OD matrix of expressing transport need ₁Matrix, OD ₂Matrix ..., OD _∏Matrix, ∏ are the positive integer greater than 1, the OD matrix after each splits comprise Γ starting and terminal point to transport need, promptly

q _i(o ₁, d ₁), q _i(o ₂, d ₂) ..., q _i(o _n, d _n) ..., q _i(o _Γ, d _Γ), q wherein _i(o _n, d _n) expression OD _iStarting point is a node in the matrix

Terminal point is a node

Transport need,

Step 2: urban road network is set at resistive directed networks G, and (V, E), (V, node E) form set of node V, are designated as: V={v as described resistive directed networks G with the terminal in crossing in the urban road network and unidirectional highway section ₁, v ₂..., v _τ, τ is the number of node, v ₁, v ₂..., v _τBe the node of directed networks, (the limit collection is designated as: E={e for V, directed edge E) as described resistive directed networks G with the unidirectional highway section in the urban road network ₁, e ₂..., e _σ..., e _η, e ₁, e ₂..., e _σ..., e _ηThe unidirectional highway section of the directed edge representative of expression directed networks, σ is the subscript in a certain unidirectional highway section, and η is the number in unidirectional highway section in the directed networks, and (V E) carries out initialization, obtains initialization directed networks G to directed networks G ₀(V, E),

Step 3: initialization impedance, the volume of traffic and the volume of traffic constitute, and make directed networks G ₀(V, E) in the accumulative total volume of traffic on all unidirectional highway sections

σ=1,2 ..., η, Represent unidirectional highway section e _σOn the accumulative total volume of traffic; Make the volume of traffic on all unidirectional highway sections constitute

σ=1,2 ..., η,

Represent unidirectional highway section e _σGo up by node v _oTo v _dThe volume of traffic, the impedance in unidirectional highway section is:

$w_{e_{\mathrm{\σ}}}=w_{e_{\mathrm{\σ}}}^0(1+\mathrm{\α}{\left(\frac{x_{e_{\mathrm{\σ}}}}{c_{e_{\mathrm{\σ}}}}\right)}^{\mathrm{\β}})$

Wherein

For when on the unidirectional highway section during without any vehicle, the time cost that vehicle passes through, it is that the geometrical length in known unidirectional highway section is divided by the known design speed of a motor vehicle;

Be the one-way road section traffic volume amount,

Be the maximum traffic capacity in unidirectional highway section,

Be unidirectional highway section saturation degree; α and β are empirical parameter, α=0.15, and β=4.0,

Step 4: loop initialization pointer i=1,

Step 5: judge i≤∏, if execution in step 6 then; Otherwise, finishing, the volume of traffic that obtains on every highway section constitutes

σ=1,2 ..., η, 1≤o≤τ, 1≤d≤τ,

Step 6: calculate current impedance matrix and be source point and with any node respectively with the impedance matrix generating algorithm, search shortest path for traffic in distributing and prepare with respect to the preposition node of another node arbitrarily,

Step 7: use the traffic allocation algorithm with transport need OD _iMatrix allocation arrives last time directed networks G _I-1(V, E) on, form current directed networks G _i(V E), and records to network G _i(V, E) volume of traffic on every the unidirectional highway section on constitutes,

Step 8: make i=i+1, return step 5,

Described impedance matrix generating algorithm is:

Step 6.1: loop initialization pointer j=1,

Step 6.2: if j≤τ, then execution in step 6.3, otherwise impedance matrix generates and to finish, and obtains the impedance between each node, and to obtain respectively with any node be source point and with respect to the preposition node of another node arbitrarily,

Step 6.3: adopt single source impedance generating algorithm to calculate arbitrary node v _jArrive all node v as source point _kImpedance r _j[v _k], k=1,2 ... τ, and obtain preposition node with respect to arbitrary node of source point,

Step 6.4: make j=j+1, return step 6.2,

Wherein, the described single source impedance generating algorithm of step 6.3 is:

Step 6.3.1: initialization p _j[] array, r _j[] array, S set, set Q,

p _j[v _k] represent with node v _jNode v for source point _kPreposition node, k=1,2 ... τ, the order:

p _j[v _k]＝φ

r _j[v _k] represent with source point v _jBe starting point, node v _kBe the path impedance of terminal point, k=1,2 ... τ, order:

$r_j\left[v_k\right]=\left\{\begin{array}{cc}0& k=j\\ \text{\∝}& k\≠j\end{array}\right.$

S set is used to deposit the node of having handled, and initialization S makes S=φ; Set Q is used to deposit untreated node, and initialization Q makes Q=V,

Step 6.3.2: judge that whether the Q set is empty, if be empty, then obtains node v _jArrive all node v as source point _kImpedance r _j[v _k], k=1,2 ..., τ, and obtain preposition node with respect to arbitrary node of source point; Otherwise, execution in step 6.3.3,

Step 6.3.3: establishing among the current set Q has Ψ node, and Ψ≤τ according to node subscript series arrangement from small to large, and corresponds respectively to l with tactic node subscript to the node among the current Q ₁, l ₂..., l _Ψ, obtain

1≤l _ξ≤ τ,

Step 6.3.4: check

Therefrom find out minimum And with node

As node u,

Step 6.3.5: determine source point v respectively _jTo the impedance of each adjacent node of node u, and definite source point v _jThe preposition node of adjacent node of node u,

Step 6.3.6: node u is moved to the S set by the Q set, execution in step 6.3.2,

Wherein, the described definite respectively source point v of step 6.3.5 _jTo the impedance of each adjacent node of node u, and definite source point v _jThe method of preposition node of adjacent node of node u be:

Step 6.3.5.1: establishing node u has Ω adjacent node, and Ω≤τ according to subscript series arrangement from small to large, and corresponds respectively to m with tactic node subscript to the adjacent node of node u ₁, m ₂..., m _Ω, obtain the adjacent node of tactic node u

$v_{{\mathrm{<figure-callout\; id="1"\; label="m"\; filenames="DEST\_PATH\_HSB00000294871800021.png,DEST\_PATH\_HSB00000294871800031.png"\; state="\{\{state\}\}">m</figure-callout>}}_1},v_{{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="DEST\_PATH\_HSB00000294871800011.png,DEST\_PATH\_HSB00000294871800051.png"\; state="\{\{state\}\}">m</figure-callout>}}_2},...,v_{m_{\mathrm{\&rho;}}},...,v_{m_{\mathrm{\&Omega;}}},1\&le;{,m}_{\mathrm{\&rho;}}\&le;\mathrm{\&tau;},$

Step 6.3.5.2: establish cycle index g=1,

Step 6.3.5.3: if g≤Ω, execution in step 6.3.5.4 then, otherwise, current source point v obtained _jTo the impedance of each adjacent node of node u, and source point v _jThe preposition node of adjacent node of node u,

Step 6.3.5.4: if

Execution in step 6.3.5.5 then; Otherwise, skip to step 6.3.5.7, wherein,

Expression by node u to adjacent node

The impedance of unidirectional highway section, if u is extremely

Between many unidirectional highway sections are arranged, then

Refer to the impedance on that unidirectional highway section of impedance minimum,

Step 6.3.5.5: make source point v _jTo adjacent node

Impedance equal source point v _jTo the impedance of u add u with

Impedance on the unidirectional highway section, even

Step 6.3.5.6: order With respect to v _jFor the preposition node of starting point is u, even

Step 6.3.5.7: make g=g+1, return step 6.3.5.3;

Step 7 is described " uses the traffic allocation algorithm with transport need OD _iMatrix allocation to directed networks G last time (V, E) on, form current directed networks G (V, E), and directed networks G before the record (V, E) volume of traffic on every the unidirectional highway section on constitutes " method be:

Step 7.1: establish transport need OD _iComprise in the matrix Γ starting and terminal point to transport need, promptly

q _i(o ₁, d ₁), q _i(o ₂, d ₂) ..., q _i(o _n, d _n) ..., q _i(o _Γ, d _Γ), q wherein _i(o _n, d _n) represent that starting point is a node in the ODi matrix

Terminal point is a node Transport need,

Step 7.2: establish cycle index n=1,

Step 7.3: if n≤Γ, then execution in step 7.4; Otherwise, obtain current one-way road section traffic volume amount and road section traffic volume and form,

Step 7.4: with starting point-terminal point to allocation algorithm distribute the starting point that obtains by step 1

Terminal point

Between transport need q _i(o _n, d _n),

Step 7.5: make n=n+1, return step 7.3,

Wherein, the starting point-terminal point described in the step 7.4 to allocation algorithm be:

Step 7.4.1: establish pointer s=d _n, d _nBe the node subscript of the terminal point of point-to-point transmission transport need,

Step 7.4.2: if pointer s=o _n, then obtain transport need q _i(o _n, d _n) in starting point be

Terminal point is

Shortest path on allocation result, otherwise, execution in step 7.4.3,

Step 7.4.3: find source point to be

Node v _sPreposition node

Node v then _sWith source point be Node v _sPreposition node v ' _sBetween the unidirectional highway section of unidirectional highway section impedance minimum be e _a,

Step 7.4.4: at the unidirectional highway section e of unidirectional highway section impedance minimum _aLast distribute traffic amount, order accumulative total one-way road section traffic volume amount

$x_{e_a}=x_{e_a}+q_i(o_n,d_n),$

Step 7.4.5: record is to the unidirectional highway section e of highway section impedance minimum _aOn the volume of traffic constitute, the order accumulative total volume of traffic constitutes

$T_{e_a}(o_n,d_n)=T_{e_a}(o_n,d_n)+q_i(o_n,d_n),$

Step 7.4.6: make pointer s=s ', return step 7.4.2 then.

3, technique effect:

(1) each volume of traffic starting point that list of missing articles distributes to the highway section in the conventional allocation method, the shortcoming of endpoint information have been overcome, the assay method that one-way road section traffic volume constitutes has write down the starting point of each volume of traffic that distributes, the point-to-point transmission of terminal point in assigning process, the traffic that forms the highway section constitutes.

(2) in the assigning process, this method has adopted the impedance matrix generating algorithm, and is more superior than traditional floyd-warshall algorithm.It is, under the prerequisite of given adjacency list, is τ with the time complexity of impedance matrix generating algorithm ₀(η+τ log τ), and the time complexity of floyd-warshall algorithm is τ ³, the impedance matrix generating algorithm is more excellent.

(3) assay method that constitutes of one-way road section traffic volume provides a kind of system of new analysis road traffic for the urban planner.In fact, soil utilization and urban transportation are the relations of presenting mutually.Traditional traffic distributes only provides the total amount of one-way road section traffic volume, and the specifying information of these traffic can not be provided.This method provides starting point, the terminal point of each traffic in the unidirectional highway section, and the volume of traffic summation that can add up identical starting point and terminal point, by the record in the assigning process, three complete information that traffic constitutes in the unidirectional highway section are provided, make the urban planner can pass through these information, adjust generation, the traffic attraction in corresponding plot targetedly.For the urban planner has solved in land use planning, only provided the traffic distribution result in the past, and do not have adjust according to the difficult problem of goal directness.

Description of drawings

Fig. 1 is urban road network's exemplary plot.

Fig. 2 is main process figure.

Fig. 3 is the impedance matrix process flow diagram.

Fig. 4 is single source impedance product process figure.

Fig. 5 is the node processing algorithm flow chart.

Fig. 6 is traffic allocation flow figure.

Fig. 7 is a point-to-point transmission allocation algorithm process flow diagram.

Fig. 8 is urban road network's exemplary plot.

Fig. 9 is the figure of initialization urban road network.

Embodiment

The assay method main body that road section traffic volume constitutes: the assay method that road section traffic volume constitutes is based upon on follow-on method of traffic assignment.This method mainly is made of four steps, comprises that traffic generation and attraction, Traffic Distribution, mode of transportation are divided and traffic distributes, and it is mainly to the quadravalence section---and the traffic distribution improves.

General thought is: " initialization demand (fractionation) " → " initialization road network " → " the initialization volume of traffic " → " handling each part transport need successively ".To each part transport need: at first set up impedance matrix according to the accumulative total volume of traffic on road network and current each highway section; Carry out traffic again and distribute, with result and the stack of the accumulative total volume of traffic; Write down road section traffic volume simultaneously and constitute, constitute stack with the accumulative total road section traffic volume, as Fig. 2.

Embodiment 1

Step 1: the initialization transport need on average is split as ∏ five equilibrium: OD with the OD matrix of expressing transport need ₁Matrix, OD ₂Matrix ..., OD _∏Matrix, ∏ are the positive integer greater than 1,

Step 2: the initialization road network, urban road network is set at resistive directed networks G, and (V, E), (V, node E) form set of node (V), are designated as: V={v as described resistive directed networks G with the terminal in crossing in the urban road network and highway section ₁, v ₂... v _τ, τ is the number of node, v ₁, v ₂... v _τBe the node of directed networks, (the limit collection is designated as: E={e for V, directed edge E) as described resistive directed networks G with the unidirectional highway section in the urban road network ₁, e ₂..., e _a... e _η, e ₁, e ₂..., e _a... e _ηThe limit of expression directed networks, a is the subscript in a certain highway section, and η is the number in highway section in the directed networks, and as Fig. 1, the calculating formula of impedance is:

$w_{e_a}=w_{e_a}^0(1+\mathrm{\&alpha;}{\left(\frac{x_{e_a}}{c_{e_a}}\right)}^{\mathrm{\&beta;}})$

Wherein For when on the highway section during without any vehicle, the time cost that vehicle passes through, it is relevant with geometrical length, the design speed in known highway section;

Be the road section traffic volume amount,

Be the maximum traffic capacity in highway section,

Be the highway section saturation degree; α and β are empirical parameter, α=0.15, and β=4.0,

Step 3: the initialization volume of traffic and the volume of traffic constitute, and make the accumulative total volume of traffic on all highway sections

Expression highway section e _aOn the accumulative total volume of traffic; Make the volume of traffic on all highway sections constitute Expression highway section e _aGo up by node v _oTo v _dThe volume of traffic,

Step 4: loop initialization pointer i=1,

Step 5: judge i≤∏, if execution in step 6 then; Otherwise, finish,

Step 6: calculate current impedance matrix with the impedance matrix generating algorithm, and arbitrary node point prepares for searching shortest path in the traffic distribution with respect to the preposition node of arbitrary node,

Step 7: at first, find out shortest path, use the traffic allocation algorithm with transport need OD by the result of step 6 _iMatrix allocation is to road network, and the volume of traffic that writes down on every highway section constitutes,

Step 8: make i=i+1, return step 5.

The impedance matrix generating algorithm

One, the generating algorithm of impedance matrix

At first, be source point with each node, calculate the minimum impedance that they arrive all nodes; Again result of calculation is compiled and formed impedance matrix.As Fig. 3.

It is as follows that impedance matrix generates step:

Step 6.1: loop initialization pointer j=1,

Step 6.2: if j≤τ, then execution in step 6.3, otherwise, impedance matrix generates and finishes, obtain the impedance between each node, and to obtain each node being the preposition node of starting point with respect to arbitrary node, is the preposition node of starting point with respect to arbitrary node by obtaining each node, we can obtain internodal shortest path at the traffic allocated phase

Step 6.3: adopt single source impedance generating algorithm to calculate arbitrary node v _jArrive all node v as source point _kImpedance r _j[v _k], k=1,2 ... τ, and to obtain arbitrary node be the preposition node of starting point with respect to source point, prepares for later traffic distributes,

Step 6.4: make j=j+1, return step 6.2.

Two, single source impedance generating algorithm

What algorithm solved is the shortest route problem of individual node in the transportation network (source point) to all nodes.

The input of algorithm has comprised a resistive transportation network G, and one of them comes source node v _j, j=1,2, ..., τ.We represent the set of all nodes among the G with V.Limit among each figure all is that two nodes are formed has order elements right.We represent the set on all limits with E, and the impedance on limit then by impedance function w definition, spends in the traffic assignment problem middle impedance w representative time.The definition of w is Bureau of Public Roads model (BPR), wherein w ₀For when on the highway section during without any vehicle, the time cost that vehicle passes through, it is relevant with geometrical length, the design speed in known highway section;

Be link flow,

Be the highway section capacity,

Be the highway section saturation degree; α and β are empirical parameter, α=0.15, β=4.0.

$w\left(e_a\right)=w_0\left(e_a\right)(1+\mathrm{\&alpha;}{\left(\frac{x_{e_a}}{c_{e_a}}\right)}^{\mathrm{\&beta;}})$

The impedance on limit is two time costs between the node.Appointing the impedance of inter-two-point path, is exactly the impedance summation in all highway sections on this path.Node v is arranged among the known V _j, algorithm can find from a node v _jTake office the what shortest path of his node.

This algorithm is by being each node v _kKeep found so far from v _jTo v _kShortest path come work.When initial, source point v _jThe path value be 0 (r by tax _j[v _j]=0), simultaneously the path r of every other node _j[v _k] be made as infinity, represent that promptly we do not know that any path of these nodes of leading to is (for all node v among the V _kRemove v _jOuter r _j[v _k]=∞); The father node of all nodes is composed and is empty (p simultaneously _j[v _k]=φ).When algorithm finishes, array r _jWhat store in [] is from v _jTo v _kThe impedance of shortest path, if perhaps the non-existent words in path are infinitely great.

Concrete steps are as follows:

Step 6.3.1: initialization p _j[] array, r _j[] array, S set, set Q,

p _j[v _k] expression v _kWith respect to v _jBe the preposition node of starting point, k=1,2 ... τ, order:

p _j[v _k]＝φ

r _j[v _k] the expression starting point is v _j, terminal point is v _kThe impedance of shortest path, k=1,2 ... τ, order:

$r_j\left[v_k\right]=\left\{\begin{array}{cc}0& k=j\\ \&Proportional;& k\&NotEqual;j\end{array}\right.$

S set is used to deposit the node of having handled, and initialization S makes S=φ; Set Q is used to deposit untreated node, and initialization Q makes Q=V,

Step 6.3.2: judge that whether the Q set is empty, if be empty, then finishes; Otherwise, execution in step 6.6.3,

Step 6.3.3: establishing among the set Q has Ψ node, Ψ≤τ.Node among the current Q is arranged from small to large according to subscript:

L wherein _ξBe the subscript of node v, 1≤l _ξ≤ τ, l _ξBe the little node subscript of node subscript ξ among the Q,

Step 6.3.4: check Therefrom find out minimum

Order

Step 6.3.5: the adjacent node of determining node u is with respect to source point v _jPreposition node and source point v for starting point _jTo the impedance of the adjacent node of u, prepare for obtaining shortest path in the traffic allocation algorithm hereinafter,

Step 6.3.6: node u is moved to S set, execution in step 6.3.2 by the Q set.

Wherein, the adjacent node of described definite node u is with respect to source point v _jPreposition node and source point v for starting point _jImpedance algorithms to the adjacent node of u is:

Step 6.3.5.1: establishing u has Ω adjacent node, Ω≤τ.Adjacent node to u is arranged from small to large according to subscript:

M wherein _gBe the subscript of node v, 1≤m _g≤ τ, m _gBe the little node subscript of subscript g in the adjacent node of u,

Step 6.3.5.2: establish cycle index g=1,

Step 6.3.5.3: if g≤Ω, execution in step 6.3.5.4 then, otherwise node processing finishes,

Step 6.3.5.4: if

Execution in step 6.3.5.5 then; Otherwise, skip to step 6.3.5.7, wherein,

Expression by node u to adjacent node

The highway section impedance, if u is extremely

Between many highway sections are arranged, then

Refer to the impedance on that highway section of impedance minimum,

Step 6.3.5.5: make source point v _jTo adjacent node

Impedance equal source point v _jTo the impedance of u add u with

Impedance on the highway section, even

$r_j\left[v_{m_g}\right]=r_j\left[u\right]+w_u\&RightArrow;v_{m_g},$

Step 6.3.5.6: order

With respect to v _jFor the preposition node of starting point is u, even

Step 6.3.5.7: make g=g+1, return step 6.3.5.3.

The traffic allocation algorithm

One, mathematical model

The apportion model commonly used of urban road network has: entirely by completely without model (All or Nothing), user equilibrium model (UserEquilibrium), polynary Probit model.Here we use the user equilibrium model to come the distribute traffic amount.The mathematic(al) representation of user equilibrium model is:

$\min z\left(x\right)=\underset{a=1}{\overset{\mathrm{\&eta;}}{\mathrm{\&Sigma;}}}{\&Integral;}_0^{x_{e_a}}{w}_{e_a}\left(t\right)\mathrm{dt}$

Two, arthmetic statement

Finding the solution of user equilibrium model usually is the Frank-Wolfe method of finding the solution the linear restriction quadratic programming by a kind of.Can obtain exact solution by this method, but the operation efficiency of this method is lower, is not suitable for engineering calculation.Therefore we adopt the increment apportion design to solve the user equilibrium problem in program.The increment distribution is meant OD demand average mark is slit into some parts, is loaded on the road network successively.The highway section can become along with the increase of the volume of traffic and block up, and the highway section also can increase by the time, and original time shortest path will change.Distribute when finishing, which bar road no matter the user of identical starting and terminal point pass through, and institute's time spent is identical, has just reached " user equilibrium ".

Concrete steps are as follows:

Step 7.1: establish transport need OD _iComprise in the matrix Γ starting and terminal point to transport need, promptly

q _i(o ₁, d ₁), q _i(o ₂, d ₂) ..., q _i(o _n, d _n) ..., q _i(o _Γ, d _Γ), q wherein _i(o _n, d _n) expression OD _iStarting point is in the matrix Terminal point is Transport need,

Step 7.2: establish cycle index n=1,

Step 7.3: if n≤Γ, then execution in step 7.4; Otherwise, distribute and finish,

Step 7.4: handle with the point-to-point transmission allocation algorithm

Between transport need q _i(o _n, d _n),

Step 7.5: make n=n+1, return step 7.3.

Wherein, the point-to-point transmission allocation algorithm is:

Step 7.4.1: establish pointer s=d _n, d _nBe the node subscript of the terminal point of point-to-point transmission transport need,

Step 7.4.2: if the node subscript s=o of pointed starting point _n, then finish, otherwise, execution in step 7.4.3,

Step 7.4.3: find node v _sPreposition node

And node v _sWith its preposition node v ' _sBetween highway section e _a,

Step 7.4.4: at highway section e _aLast distribute traffic amount, order accumulative total road section traffic volume amount

Step 7.4.5: record highway section e _aOn the volume of traffic constitute, the order accumulative total volume of traffic constitutes

Step 7.4.6: make the subscript of its preposition node of pointer s sensing, even s=s ' returns step 7.4.2 then.

Embodiment 2

The 1st step: suppose that transport need OD matrix is

Table 1.OD matrix

O\D?	v 1	v 2	v 3	v 4
					v 1	0?	0?	100?	0?
v 2	0?	0?	0?	50?
					v 3	0?	0?	0?	0?
v 4	0?	0?	0?	0?

It is split as 2 equal OD matrix: OD ₁, OD ₂,

OD matrix after table 2. splits

O\D?	v 1	v 2	v 3	v 4
					v 1	0?	0?	50?	0?
v 2	0?	0?	0?	25?
					v 3	0?	0?	0?	0?
v 4	0?	0?	0?	0?

OD then ₁={ q ₁(1,3), q ₁(2,4) }, q wherein ₁(1,3)=50, q ₁(2,4)=25,

OD ₂={ q ₂(1,3), q ₂(2,4) }, q wherein ₂(1,3)=50, q ₂(2,4)=25,

The 2nd step: suppose urban road network such as Fig. 8,

Set of node V={v ₁, v ₂, v ₃, v ₄, node number τ=4, limit collection E={e ₁, e ₂, e ₃, e ₄Represent the unidirectional highway section of each bar in the road network, and limit number η=8, the one-way road section traffic volume capacity of establishing every limit is 60.

The 3rd step: make the one-way road section traffic volume amount

One-way road section traffic volume is formed

Suppose that the zero flow impedance on the unidirectional highway section is

Because the volume of traffic on the unidirectional highway section is 0, so

As Fig. 9,

The 4th step: loop initialization pointer i=1,

The 5th step: judge i≤∏, this moment i=1, ∏=2, the judgement formula is 1≤2, correct judgment continues to carry out,

The 6th step: generate impedance matrix, loop initialization pointer j=1 at first,

The 7th step: judge j≤τ, this moment j=1, τ=4, the judgement formula is 1≤4, correct judgment continues to carry out,

The 8th step: use single source impedance method of formation to calculate the impedance r of v1 to all nodes ₁[v _k] and with respect to v ₁Preposition node,

The 9th step: initialization p ₁[] array, r ₁[] array, S set, set Q,

The 10th step: p ₁[v ₁]=φ, p ₁[v ₂]=φ, p ₁[v ₃]=φ, p ₁[v ₄]=φ,

The 11st step: r ₁[v ₁]=0, r ₁[v ₂]=∞, r ₁[v ₃]=∞, r ₁[v ₄]=∞,

The 12nd step: S=φ, Q={v ₁, v ₂, v ₃, v ₄,

The 13rd step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, v ₄, 4 nodes are arranged among the Q, Ψ=4 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3, l ₄=4,

The 14th step: check r ₁[v ₁], r ₁[v ₂], r ₁[v ₃], r ₁[v ₄], be respectively 0.0, ∞, ∞, so ∞ is l _λ=1, make u=v ₁,

The 15th step: establishing node u has Ω adjacent node, u=v ₁, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 16th step: establish cycle index g=1,

The 17th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 18th step: judge

This moment j=1, g=1, mg=2, r _j[m _g]=∞, u=v ₁, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞1.0, and correct judgment then continues to carry out,

The 19th step: order

Be r ₁[v ₂]=0.0+1.0=1.0,

The 20th step: order

Be p ₁[v ₂]=v ₁,

The 21st step: make g=g+1, thus g=2,

The 22nd step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 23rd step: judge

This moment j=1, g=2, mg=4, r _j[m _g]=∞, u=v ₁, r _j[u]=0.0, Therefore, the judgement formula is ∞＞2.0, and correct judgment then continues to carry out,

The 24th step: order

Be r ₁[v ₄]=0.0+2.0=2.0,

The 25th step: order

Be p ₁[v ₄]=v ₁,

The 26th step: make g=g+1, thus g=3,

The 27th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 28th step: u is moved to S set, u=v by the Q set ₁,

The 29th step: judge whether current set Q is empty, at this moment Q={v ₂, v ₃, v ₄, 3 nodes are arranged among the Q, Ψ=3 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=2, l ₂=3, l ₃=4,

The 30th step: check r ₁[v ₂], r ₁[v ₃], r ₁[v ₄], be respectively 1.0, ∞, 2.0, so l _λ=2, make u=v ₂,

The 31st step: establishing node u has Ω adjacent node, u=v ₂, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 32nd step: establish cycle index g=1,

The 33rd step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 34th step: judge

This moment j=1, g=1, m _g=1, r _j[m _g]=0.0, u=v ₂, r _j[u]=1.0,

Therefore, the judgement formula is 0.0＞2.0, and next adjacent node is judged in misjudgment

The 35th step: make g=g+1, thus g=2,

The 36th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 37th step: judge

This moment j=1, g=2, m _g=3, r _j[m _g]=∞, u=v ₂, r _j[u]=1.0,

Therefore, the judgement formula is ∞＞4.0, and correct judgment then continues to carry out,

The 38th step: order

Be r ₁[v ₃]=0.0+3.0=3.0,

The 39th step: order Be p ₁[v ₃]=v ₂,

The 40th step: make g=g+1, thus g=3,

The 41st step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 42nd step: u is moved to S set, u=v by the Q set ₂,

The 43rd step: judge whether current set Q is empty, at this moment Q={v ₃, v ₄, 2 nodes are arranged among the Q, Ψ=2 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=3, l ₂=4,

The 44th step: check r ₁[v ₃], r ₁[v ₄], be respectively 4.0,2.0, so l _λ=4, make u=v ₄,

The 45th step: establishing node u has Ω adjacent node, u=v ₄, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 46th step: establish cycle index g=1,

The 47th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 48th step: judge

This moment j=1, g=1, m _g=1, r _j[m _g]=0.0, u=v ₄, r _j[u]=2.0, Therefore, the judgement formula is 0.0＞4.0, and next adjacent node is judged in misjudgment

The 49th step: make g=g+1, thus g=2,

The 50th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 51st step: judge

This moment j=1, g=2, m _g=3, r _j[m _g]=4.0, u=v ₄, r _j[u]=2.0,

Therefore, the judgement formula is 4.0＞3.0, and correct judgment then continues to carry out,

The 52nd step: order

Be r ₁[v ₃]=0.0+1.0=1.0,

The 53rd step: order

Be p ₁[v ₃]=v ₄,

The 54th step: make g=g+1, thus g=3,

The 55th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 56th step: u is moved to S set, u=v by the Q set ₄,

The 57th step: judge whether current set Q is empty, at this moment Q={v ₃, 1 node is arranged among the Q, Ψ=1 according to node subscript series arrangement from small to large, obtains l to the node among the current Q ₁=3,

The 58th step: check r ₁[v ₃], be respectively 3.0, so l _λ=3, make u=v ₃,

The 59th step: establishing node u has Ω adjacent node, u=v ₃, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 60th step: establish cycle index g=1,

The 61st step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 62nd step: judge

This moment j=1, g=1, m _g=2, r _j[m _g]=1.0, u=v ₃, r _j[u]=3.0,

Therefore, the judgement formula is 1.0＞6.0, and next adjacent node is judged in misjudgment

The 63rd step: make g=g+1, thus g=2,

The 64th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 65th step: judge

This moment j=1, g=2, m _g=4, r _j[m _g]=2.0, u=v ₃, r _j[u]=3.0,

Therefore, the judgement formula is 2.0＞4.0, and next adjacent node is judged in misjudgment

The 66th step: make g=g+1, thus g=3,

The 67th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 68th step: u is moved to S set, u=v by the Q set ₃,

The 69th step: judge whether current set Q is empty, and this moment, Q=φ then jumped out circulation,

The 70th step: judge j≤τ, this moment j=2, τ=4, the judgement formula is 2≤4, correct judgment continues to carry out,

The 71st step: use single source impedance method of formation to calculate v ₂Impedance r to all nodes ₂[v _k] and with respect to v ₂Preposition node,

The 72nd step: initialization p ₂[] array, r ₂[] array, S set, set Q,

The 73rd step: p ₂[v ₁]=φ, p ₂[v ₂]=φ, p ₂[v ₃]=φ, p ₂[v ₄]=φ,

The 74th step: r ₂[v ₁]=∞, r ₂[v ₂]=0, r ₂[v ₃]=∞, r ₂[v ₄]=∞,

The 75th step: S=φ, Q=[v ₁, v ₂, v ₃, v ₄,

The 76th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, v ₄, 4 nodes are arranged among the Q, Ψ=4 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3, l ₄=4,

The 77th step: check r ₂[v ₁], r ₂[v ₂], r ₂[v ₃], r ₂[v ₄], be respectively ∞, 0.0, ∞, so ∞ is l _λ=2, make u=v ₂,

The 78th step: establishing node u has Ω adjacent node, u=v ₂, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript from little not big series arrangement, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 79th step: establish cycle index g=1,

The 80th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 81st step: judge

This moment j=2, g=1, m _g=1, r _j[m _g]=∞, u=v ₂, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞1.0, and correct judgment then continues to carry out,

The 82nd step: order

Be r ₂[v ₁]=0.0+1.0=1.0,

The 83rd step: order

Be p ₂[v ₁]=v ₂,

The 84th step: make g=g+1, thus g=2,

The 85th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 86th step: judge

This moment j=2, g=2, m _g=3, r _j[m _g]=∞, u=v ₂, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞3.0, and correct judgment then continues to carry out,

The 87th step: order

Be r ₂[v ₃]=0.0+3.0=3.0,

The 88th step: order

Be p ₂[v ₃]=v ₂,

The 89th step: make g=g+1, thus g=3,

The 90th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 91st step: u is moved to S set, u=v by the Q set ₂,

The 92nd step: judge whether current set Q is empty, at this moment Q={v ₁, v ₃, v ₄, 3 nodes are arranged among the Q, Ψ=3 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=3, l ₃=4,

The 93rd step: check r ₂[v ₁], r ₂[v ₃], r ₂[v ₄], being respectively 1.0,3.0, so ∞ is l _λ=1, make u=v ₁,

The 94th step: establishing node u has Ω adjacent node, u=v ₁, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 95th step: establish cycle index g=1,

The 96th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 97th step: judge

This moment j=2, g=1, m _g=2, r _j[m _g]=0.0, u=v ₁, r _j[u]=1.0,

Therefore, the judgement formula is 0.0＞2.0, and next adjacent node is judged in misjudgment

The 98th step: make g=g+1, thus g=2,

The 99th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 100th step: judge

This moment j=2, g=2, m _g=4, r _j[m _g]=∞, u=v ₁, r _j[u]=1.0,

Therefore, the judgement formula is ∞＞3.0, and correct judgment then continues to carry out,

The 101st step: order

Be r ₂[v ₄]=0.0+2.0=2.0,

The 102nd step: order

Be p ₂[v ₄]=v ₁,

The 103rd step: make g=g+1, thus g=3,

The 104th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then looked into the distance out in misjudgment,

The 105th step: u is moved to S set, u=v by the Q set ₁,

The 106th step: judge whether current set Q is empty, at this moment Q={v ₃, v ₄, 2 nodes are arranged among the Q, Ψ=2 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=3, l ₂=4,

The 107th step: check r ₂[v ₃], r ₂[v ₄], be respectively 3.0,3.0, so l _λ=3, make u=v ₃,

The 108th step: establishing node u has Ω adjacent node, u=v ₃, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 109th step: establish cycle index g=1,

The 110th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 111st step: judge This moment j=2, g=1, m _g=2, r _j[m _g]=0.0, u=v ₃, r _j[u]=3.0,

Therefore, the judgement formula is 0.0＞6.0, and next adjacent node is judged in misjudgment

The 112nd step: make g=g+1, thus g=2,

The 113rd step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 114th step: judge

This moment j=2, g=2, m _g=4, r _j[m _g]=3.0, u=v ₃, r _j[u]=3.0,

Therefore, the judgement formula is 3.0＞4.0, and next adjacent node is judged in misjudgment

The 115th step: make g=g+1, thus g=3,

The 116th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 117th step: u is moved to S set, u=v by the Q set ₃,

The 118th step: judge whether current set Q is empty, at this moment Q={v ₄, 1 node is arranged among the Q, Ψ=1 according to node subscript series arrangement from small to large, obtains l to the node among the current Q ₁=4,

The 119th step: check r ₂[v ₄], be respectively 3.0, so l _λ=4, make u=v ₄,

The 120th step: establishing node u has Ω adjacent node, u=v ₄, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 121st step: establish cycle index g=1,

The 122nd step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 123rd step: judge

This moment j=2, g=1, m _g=1, r _j[m _g]=1.0, u=v ₄, r _j[u]=3.0,

Therefore, the judgement formula is 1.0＞5.0, and next adjacent node is judged in misjudgment

The 124th step: make g=g+1, thus g=2,

The 125th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 126th step: judge

This moment j=2, g=2, m _g=3, r _j[m _g]=3.0, u=v ₄, r _j[u]=3.0, Therefore, the judgement formula is 3.0＞4.0, and next adjacent node is judged in misjudgment

The 127th step: make g=g+1, thus g=3,

The 128th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 129th step: u is moved to S set, u=v by the Q set ₄,

The 130th step: judge whether current set Q is empty, and this moment, Q=φ then jumped out circulation,

The 131st step: judge j≤τ, this moment j=3, τ=4, the judgement formula is 3≤4, correct judgment continues to carry out,

The 132nd step: use single source impedance method of formation to calculate v ₃Impedance r to all nodes ₃[v _k] and with respect to v ₃Preposition node,

The 133rd step: initialization p ₃[] array, r ₃[] array, S set, set Q,

The 134th step: p ₃[v ₁]=φ, p ₃[v ₂] a=φ, p ₃[v ₃]=φ, p ₃[v ₄]=φ,

The 135th step: r ₃[v ₁]=∞, r ₃[v ₂]=∞, r ₃[v ₃]=0, r ₃[v ₄]=∞,

The 136th step: S=φ, Q={v ₁, v ₂, v ₃, v ₄,

The 137th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, v ₄, 4 nodes are arranged among the Q, Ψ=4 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3, l ₄=4,

The 138th step: check r ₃[v ₁], r ₃[v ₂], r ₃[v ₃], r ₃[v ₄], be respectively ∞, ∞, 0.0, so ∞ is l _λ=3, make u=v ₃,

The 139th step: establishing node u has Ω adjacent node, u=v ₃, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 140th step: establish cycle index g=1,

The 141st step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 142nd step: judge

This moment j=3, g=1, m _g=2, r _j[m _g]=∞, u=v ₃, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞3.0, and correct judgment then continues to carry out,

The 143rd step: order

Be r ₃[v ₂]=0.0+3.0=3.0,

The 144th step: order

Be p ₃[v ₂]=v ₃,

The 145th step: make g=g+1, thus g=2,

The 146th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 147th step: judge

This moment j=3, g=2, m _g=4, r _j[m _g]=∞, u=v ₃, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞1.0, and correct judgment then continues to carry out,

The 148th step: order

Be r ₃[v ₄]=0.0+1.0=1.0,

The 149th step: order Be p ₃[v ₄]=v ₃,

The 150th step: make g=g+1, thus g=3,

The 151st step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 152nd step: u is moved to S set, u=v by the Q set ₃,

The 153rd step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₄, 3 nodes are arranged among the Q, Ψ=3 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=4,

The 154th step: check r ₃[v ₁], r ₃[v ₂], r ₃[v ₄], be respectively ∞, 3.0,1.0, so l _λ=4, make u=v ₄,

The 155th step: establishing node u has Ω adjacent node, u=v ₄, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 156th step: establish cycle index g=1,

The 157th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 158th step: judge This moment j=3, g=1, m _g=1, r _j[m _g]=∞, u=v ₄, r _j[u]=1.0, Therefore, the judgement formula is ∞＞3.0, and correct judgment then continues to carry out,

The 159th step: order

Be r ₃[v ₁]=0.0+2.0=2.0,

The 160th step: order

Be p ₃[v ₁]=v ₄,

The 161st step: make g=g+1, thus g=2,

The 162nd step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 163rd step: judge This moment j=3, g=2, m _g=3, r _j[m _g]=0.0, u=v ₄, r _j[u]=1.0,

Therefore, the judgement formula is 0.0＞2.0, and next adjacent node is judged in misjudgment

The 164th step: make g=g+1, thus g=3,

The 165th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 166th step: u is moved to S set, u=v by the Q set ₄,

The 167th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, 2 nodes are arranged among the Q, Ψ=2 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2,

The 168th step: check r ₃[v ₁], r ₃[v ₂], be respectively 3.0,3.0, so l _λ=1, make u=v ₁,

The 169th step: establishing node u has Ω adjacent node, u=v ₁, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 170th step: establish cycle index g=1,

The 171st step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 172nd step: judge

This moment j=3, g=1, m _g=2, r _j[m _g]=3.0, u=v ₁, r _j[u]=3.0, Therefore, the judgement formula is 3.0＞4.0, and next adjacent node is judged in misjudgment

The 173rd step: make g=g+1, thus g=2,

The 174th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 175th step: judge

This moment j=3, g=2, m _g=4, r _j[m _g]=1.0, u=v ₁, r _j[u]=3.0,

Therefore, the judgement formula is 1.0＞5.0, and next adjacent node is judged in misjudgment

The 176th step: make g=g+1, thus g=3,

The 177th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 178th step: u is moved to S set, u=v by the Q set ₁,

The 179th step: judge whether current set Q is empty, at this moment Q={v ₂, 1 node is arranged among the Q, Ψ=1 according to node subscript series arrangement from small to large, obtains l to the node among the current Q ₁=2,

The 180th step: check r ₃[v ₂], be respectively 3.0, so l _λ=2, make u=u ₂,

The 181st step: establishing node u has Ω adjacent node, u=v ₂, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 182nd step: establish cycle index g=1,

The 183rd step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 184th step: judge

This moment j=3, g=1, m _g=1, r _j[m _g]=3.0, u=v ₂, r _j[u]=3.0, Therefore, the judgement formula is 3.0＞4.0, and next adjacent node is judged in misjudgment

The 185th step: make g=g+1, thus g=2,

The 186th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 187th step: judge

This moment j=3, g=2, m _g=3, r _j[m _g]=0.0, u=v ₂, r _j[u]=3.0, Therefore, the judgement formula is 0.0＞6.0, and next adjacent node is judged in misjudgment

The 188th step: make g=g+1, thus g=3,

The 189th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 190th step: u is moved to S set, u=v by the Q set ₂,

The 191st step: judge whether current set Q is empty, and this moment, Q=φ then jumped out circulation,

The 192nd step: judge j≤τ, this moment j=4, τ=4, the judgement formula is 4≤4, correct judgment continues to carry out,

The 193rd step: use single source impedance method of formation to calculate v ₄Impedance r to all nodes ₄[v _k] and with respect to v ₄Preposition node,

The 194th step: initialization p ₄[] array, r ₄[] array, S set, set Q,

The 195th step: p ₄[v ₁]=φ, p ₄[v ₂]=φ, p ₄[v ₃]=φ, p ₄[v ₄]=φ,

The 196th step: r ₄[v ₁]=∞, r ₄[v ₂]=∞, r ₄[v ₃]=∞, r ₄[v ₄]=0,

The 197th step: S=φ, Q={v ₁, v ₂, v ₃, v ₄,

The 198th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, v ₄, 4 nodes are arranged among the Q, Ψ=4 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3, l ₄=4,

The 199th step: check r ₄[v ₁], r ₄[v ₂], r ₄[v ₃], r ₄[v ₄], be respectively ∞, ∞, ∞, 0.0, so l _λ=4, make u=v ₄,

The 200th step: establishing node u has Ω adjacent node, u=v ₄, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 201st step: establish cycle index g=1,

The 202nd step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 203rd step: judge

This moment j=4, g=1, m _g=1, r _j[m _g]=∞, u=v ₄, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞2.0, and correct judgment then continues to carry out,

The 204th step: order

Be r ₄[v ₁]=0.0+2.0=2.0,

The 205th step: order

Be p ₄[v ₁]=v ₄,

The 206th step: make g=g+1, thus g=2,

The 207th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 208th step: judge

This moment j=4, g=2, m _g=3, r _j[m _g]=∞, u=v ₄, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞1.0, and correct judgment then continues to carry out,

The 209th step: order

Be r ₄[v ₃]=0.0+1.0=1.0,

The 210th step: order

Be p ₄[v ₃]=v ₄,

The 211st step: make g=g+1, thus g=3,

The 212nd step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 213rd step: u is moved to S set, u=v by the Q set ₄,

The 214th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, 3 nodes are arranged among the Q, Ψ=3 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3,

The 215th step: check r ₄[v ₁], r ₄[v ₂] a, r ₄[v ₃], be respectively 2.0, ∞, 1.0, so l _λ=3, make u=v ₃,

The 216th step: establishing node u has Ω adjacent node, u=v ₃, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 217th step: establish cycle index g=1,

The 218th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 219th step: judge

This moment j=4, g=1, m _g=2, r _j[m _g]=∞, u=v ₃, r _j[u]=1.0,

Therefore, the judgement formula is ∞＞4.0, and correct judgment then continues to carry out,

The 220th step: order

Be r ₄[v ₂]=0.0+3.0=3.0,

The 221st step: order

Be p ₄[v ₂]=v ₃,

The 222nd step: make g=g+1, thus g=2,

The 223rd step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 224th step: judge

This moment j=4, g=2, m _g=4, r _j[m _g]=0.0, u=v ₃, r _j[u]=1.0,

Therefore, the judgement formula is 0.0＞2.0, and next adjacent node is judged in misjudgment

The 225th step: make g=g+1, thus g=3,

The 226th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 227th step: u is moved to S set, u=v by the Q set ₃,

The 228th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, 2 nodes are arranged among the Q, Ψ=2 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2,

The 229th step: check r ₄[v ₁], r ₄[v ₂], be respectively 2.0,4.0, so l _λ=1, make u=v ₁,

The 230th step: establishing node u has Ω adjacent node, u=v ₁, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 231st step: establish cycle index g=1,

The 232nd step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 233rd step: judge

This moment j=4, g=1, m _g=2, r _j[m _g]=4.0, u=v ₁, r _j[u]=2.0,

Therefore, the judgement formula is 4.0＞3.0, and correct judgment then continues to carry out,

The 234th step: order

Be r ₄[v ₂]=0.0+1.0=1.0,

The 235th step: order Be p ₄[v ₂]=v ₁,

The 236th step: make g=g+1, thus g=2,

The 237th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 238th step: judge

This moment j=4, g=2, m _g=4, r _j[m _g]=0.0, u=v ₁, r _j[u]=2.0,

Therefore, the judgement formula is 0.0＞4.0, and next adjacent node is judged in misjudgment

The 239th step: make g=g+1, thus g=3,

The 240th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 241st step: u is moved to S set, u=v by the Q set ₁,

The 242nd step: judge whether current set Q is empty, at this moment Q={v ₂, 1 node is arranged among the Q, Ψ=1 according to node subscript series arrangement from small to large, obtains l to the node among the current Q ₁=2,

The 243rd step: check r ₄[v ₂], be respectively 3.0, so l _λ=2, make u=v ₂,

The 244th step: establishing node u has Ω adjacent node, u=v ₂, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 245th step: establish cycle index g=1,

The 246th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 247th step: judge

This moment j=4, g=1, m _g=1, r _j[m _g]=2.0, u=v ₂, r _j[u]=3.0,

Therefore, the judgement formula is 2.0＞4.0, and next adjacent node is judged in misjudgment

The 248th step: make g=g+1, thus g=2,

The 249th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 250th step: judge

This moment j=4, g=2, m _g=3, r _j[m _g]=1.0, u=v ₂, r _j[u]=3.0,

Therefore, the judgement formula is 1.0＞6.0, and next adjacent node is judged in misjudgment

The 251st step: make g=g+1, thus g=3,

The 252nd step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 253rd step: u is moved to S set, u=v by the Q set ₂,

The 254th step: judge whether current set Q is empty, and this moment, Q=φ then jumped out circulation,

The 255th step: establish cycle index n=1,

The 256th step: judge n≤Γ, this moment n=1, Γ=2, correct judgment then continues to carry out,

The 257th step: establish pointer s=d _n, n=1, d _n=3, so s=3

The 258th step: judge s=o _n, this moment n=1, s=3, o _n=1, misjudgment then continues to carry out,

The 259th step: finding source point is v ₁Node v ₃Preposition node p ₁[v ₃]=v ₄, node v then ₃With node v ₄Between the unidirectional highway section of unidirectional highway section impedance minimum be e ₆,

The 260th step: at the unidirectional highway section e of unidirectional highway section impedance minimum ₆Last distribute traffic amount, order accumulative total one-way road section traffic volume amount

$x_{{\mathrm{<figure-callout\; id="6"\; label="e"\; filenames="DEST\_PATH\_HSB00000294871800011.png"\; state="\{\{state\}\}">e</figure-callout>}}_6}={\mathrm{<figure-callout\; id="6"\; label="x\; e"\; filenames="DEST\_PATH\_HSB00000294871800011.png"\; state="\{\{state\}\}">x</figure-callout>}}_{e_{}}+100.00=100.00,$

The 261st step: record is to the unidirectional highway section e of highway section impedance minimum ₆On the volume of traffic constitute, the order accumulative total volume of traffic constitutes

$T_{e_6}\left(\mathrm{1,3}\right)=T_{e_6}\left(\mathrm{1,3}\right)+100.00=100.00,$

The 262nd step: make pointer s=s '=4,

The 263rd step: judge s=o _n, this moment n=1, s=4, o _n=1, misjudgment then continues to carry out,

The 264th step: finding source point is v ₁Node v ₄Preposition node p ₁[v ₄]=v ₁, node v then ₄With node v ₁Between the unidirectional highway section of unidirectional highway section impedance minimum be e ₈,

The 265th step: at the unidirectional highway section e of unidirectional highway section impedance minimum ₈Last distribute traffic amount, order accumulative total one-way road section traffic volume amount

$x_{{\mathrm{<figure-callout\; id="8"\; label="e"\; filenames="DEST\_PATH\_HSB00000294871800011.png,DEST\_PATH\_HSB00000294871800041.png"\; state="\{\{state\}\}">e</figure-callout>}}_8}={\mathrm{<figure-callout\; id="8"\; label="x\; e"\; filenames="DEST\_PATH\_HSB00000294871800011.png,DEST\_PATH\_HSB00000294871800041.png"\; state="\{\{state\}\}">x</figure-callout>}}_{e_{}}+100.00=100.00,$

The 266th step: record is to the unidirectional highway section e of highway section impedance minimum ₈On the volume of traffic constitute, the order accumulative total volume of traffic constitutes

$T_{e_8}\left(\mathrm{1,3}\right)=T_{e_8}\left(\mathrm{1,3}\right)+100.00=100.00,$

The 267th step: make pointer s=s '=1,

The 268th step: judge s=o _n, this moment n=1, s=1, o _n=1, correct judgment is then jumped out circulation,

The 269th step: make n=n+1, thus n=2,

The 270th step: judge n≤Γ, this moment n=2, Γ=2, correct judgment then continues to carry out,

The 271st step: establish pointer s=d _n, n=2, d _n=4, so s=4

The 272nd step: judge s=o _n, this moment n=2, s=4, o _n=2, misjudgment then continues to carry out,

The 273rd step: finding source point is v ₂Node v ₄Preposition node p ₂[v ₄]=v ₁, node v then ₄With node v ₁Between the unidirectional highway section of unidirectional highway section impedance minimum be e ₈,

The 274th step: at the unidirectional highway section e of unidirectional highway section impedance minimum ₈Last distribute traffic amount, order accumulative total one-way road section traffic volume amount

$x_{{\mathrm{<figure-callout\; id="8"\; label="e"\; filenames="DEST\_PATH\_HSB00000294871800011.png,DEST\_PATH\_HSB00000294871800041.png"\; state="\{\{state\}\}">e</figure-callout>}}_8}={\mathrm{<figure-callout\; id="8"\; label="x\; e"\; filenames="DEST\_PATH\_HSB00000294871800011.png,DEST\_PATH\_HSB00000294871800041.png"\; state="\{\{state\}\}">x</figure-callout>}}_{e_{}}+50.00=150.00,$

The 275th step: record is to the unidirectional highway section e of highway section impedance minimum ₈On the volume of traffic constitute, the order accumulative total volume of traffic constitutes

$T_{e_8}\left(\mathrm{2,4}\right)=T_{e_8}\left(\mathrm{2,4}\right)+50.00=50.00,$

The 276th step: make pointer s=s '=1,

The 277th step: judge s=o _n, this moment n=2, s=1, o _n=2, misjudgment then continues to carry out,

The 278th step: finding source point is v ₂Node v ₁Preposition node p ₂[v ₁]=v ₂, node v then ₁With node v ₂Between the unidirectional highway section of unidirectional highway section impedance minimum be e ₂,

The 279th step: at the unidirectional highway section e of unidirectional highway section impedance minimum ₂Last distribute traffic amount, order accumulative total one-way road section traffic volume amount

$x_{{\mathrm{<figure-callout\; id="2"\; label="e"\; filenames="DEST\_PATH\_HSB00000294871800011.png,DEST\_PATH\_HSB00000294871800051.png"\; state="\{\{state\}\}">e</figure-callout>}}_2}={\mathrm{<figure-callout\; id="2"\; label="x\; e"\; filenames="DEST\_PATH\_HSB00000294871800011.png,DEST\_PATH\_HSB00000294871800051.png"\; state="\{\{state\}\}">x</figure-callout>}}_{e_{}}+50.00=50.00,$

The 280th step: record is to the unidirectional highway section e of highway section impedance minimum ₂On the volume of traffic constitute, the order accumulative total volume of traffic constitutes

$T_{e_2}\left(\mathrm{2,4}\right)=T_{e_2}\left(\mathrm{2,4}\right)+50.00=50.00,$

The 281st step: make pointer s=s '=2,

The 282nd step: judge s=o _n, this moment n=2, s=2, o _n=2, correct judgment is then jumped out circulation,

The 283rd step: make n=n+1, thus n=3,

The 284th step: judge n≤Γ, this moment n=3, Γ=2, circulation is then jumped out in misjudgment,

The 285th step: judge i≤∏, this moment i=2, ∏=2, the judgement formula is 2≤2, correct judgment continues to carry out,

The 286th step: generate impedance matrix, loop initialization pointer j=1 at first,

The 287th step: judge j≤τ, this moment j=1, τ=4, the judgement formula is 1≤4, correct judgment continues to carry out,

The 288th step: use single source impedance method of formation to calculate v ₁Impedance v to all nodes ₁[v _k] and with respect to v ₁Preposition node,

The 289th step: initialization p ₁[] array, r ₁[] array, S set, set Q,

The 290th step: p ₁[v ₁]=φ, p ₁[v ₂]=φ, p ₁[v ₃]=φ, p ₁[v ₄]=φ,

The 291st step: r ₁[v ₁]=0, r ₁[v ₂]=∞, r ₁[v ₃]=∞, r ₁[v ₄]=∞,

The 292nd step: S=φ, Q={v ₁, v ₂, v ₃, v ₄,

The 293rd step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, v ₄, 4 nodes are arranged among the Q, Ψ=4 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3, l ₄=4,

The 294th step: check r ₁[v ₁], r ₁[v ₂], r ₁[v ₃], r ₁[v ₄], be respectively 0.0, ∞, ∞, so ∞ is l _λ=1, make u=v ₁,

The 295th step: establishing node u has Ω adjacent node, u=v ₁, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 296th step: establish cycle index g=1,

The 297th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 298th step: judge

This moment j=1, g=1, m _g=2, r _j[m _g]=∞, u=v ₁, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞1.0, and correct judgment then continues to carry out,

The 299th step: order

Be r ₁[v ₂]=0.0+1.0=1.0,

The 300th step: order

Be p ₁[v ₂]=v ₁,

The 301st step: make g=g+1, thus g=2,

The 302nd step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 303rd step: judge

This moment j=1, g=2, m _g=4, r _j[m _g]=∞, u=v ₁, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞13.7, and correct judgment then continues to carry out,

The 304th step: order

Be r ₁[v ₄]=0.0+13.7=13.7,

The 305th step: order

Be p ₁[v ₄]=v ₁,

The 306th step: make g=g+1, thus g=3,

The 307th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 308th step: u is moved to S set, u=v by the Q set ₁,

The 309th step: judge whether current set Q is empty, at this moment Q={v ₂, v ₃, v ₄, 3 nodes are arranged among the Q, Ψ=3 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=2, l ₂=3, l ₃=4,

The 310th step: check r ₁[v ₂], r ₁[v ₃], r ₁[v ₄], be respectively 1.0, ∞, 13.7, so l _λ=2, make u=v ₂,

The 311st step: establishing node u has Ω adjacent node, u=v ₂, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 312nd step: establish cycle index g=1,

The 313rd step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 314th step: judge

This moment j=1, g=1, m _g=1, r _j[m _g]=0.0, u=v ₂, r _j[u]=1.0,

Therefore, the judgement formula is 0.0＞2.1, and next adjacent node is judged in misjudgment

The 315th step: make g=g+1, thus g=2,

The 316th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 317th step: judge This moment j=1, g=2, m _g=3, r _j[m _g]=∞, u=v ₂, r _j[u]=1.0,

Therefore, the judgement formula is ∞＞4.0, and correct judgment then continues to carry out,

The 318th step: order

Be r ₁[v ₃]=0.0+3.0=3.0,

The 319th step: order Be p ₁[v ₃]=v ₂,

The 320th step: make g=g+1, thus g=3,

The 321st step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 322nd step: u is moved to S set, u=v by the Q set ₂,

The 323rd step: judge whether current set Q is empty, at this moment Q={v ₃, v ₄, 2 nodes are arranged among the Q, Ψ=2 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=3, l ₂=4,

The 324th step: check r ₁[v ₃], r ₁[v ₄], be respectively 4.0,13.7, so l _λ=3, make u=v ₃,

The 325th step: establishing node u has Ω adjacent node, u=v ₃, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 326th step: establish cycle index g=1,

The 327th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 328th step: judge

This moment j=1, g=1, m _g=2, r _j[m _g]=1.0, u=v ₃, r _j[u]=4.0,

Therefore, the judgement formula is 1.0＞7.0, and next adjacent node is judged in misjudgment

The 329th step: make g=g+1, thus g=2,

The 330th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 331st step: judge This moment j=1, g=2, m _g=4, r _j[m _g]=13.7, u=v ₃, r _j[u]=4.0,

Therefore, the judgement formula is 13.7＞5.0, and correct judgment then continues to carry out,

The 332nd step: order

Be r ₁[v ₄]=0.0+1.0=1.0,

The 333rd step: order

Be p ₁[v ₄]=v ₃,

The 334th step: make g=g+1, thus g=3,

The 335th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 336th step: u is moved to S set, u=v by the Q set ₃,

The 337th step: judge whether current set Q is empty, at this moment Q={v ₄, 1 node is arranged among the Q, Ψ=1 according to node subscript series arrangement from small to large, obtains l to the node among the current Q ₁=4,

The 338th step: check r ₁[v ₄], be respectively 5.0, so l _λ=4, make u=v ₄,

The 339th step: establishing node u has Ω adjacent node, u=v ₄, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 340th step: establish cycle index g=1,

The 341st step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 342nd step: judge

This moment j=1, g=1, m _g=1, r _j[m _g]=0.0, u=v ₄, v _j[u]=5.0,

Therefore, the judgement formula is 0.0＞7.0, and next adjacent node is judged in misjudgment

The 343rd step: make g=g+1, thus g=2,

The 344th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 345th step: judge

This moment j=1, g=2, m _g=3, r _j[m _g]=4.0, u=v ₄, r _j[u]=5.0, Therefore, the judgement formula is 4.0＞7.2, and next adjacent node is judged in misjudgment

The 346th step: make g=g+1, thus g=3,

The 347th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 348th step: u is moved to S set, u=v by the Q set ₄,

The 349th step: judge whether current set Q is empty, and this moment, Q=φ then jumped out circulation,

The 350th step: judge j≤τ, this moment j=2, τ=4, the judgement formula is 2≤4, correct judgment continues to carry out,

The 351st step: use single source impedance method of formation to calculate v ₂Impedance r to all nodes ₂[v _k] and with respect to v ₂Preposition node,

The 352nd step: initialization p ₂[] array, r ₂[] array, S set, set Q,

The 353rd step: p ₂[v ₁]=φ, p ₂[v ₂]=φ, p ₂[v ₃]=φ, p ₂[v ₄]=φ,

The 354th step: r ₂[v ₁]=∞, r ₂[v ₂]=0, r ₂[v ₃]=∞, r ₂[v ₄]=∞,

The 355th step: S=φ, Q={v ₁, v ₂, v ₃, v ₄,

The 356th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, v ₄, 4 nodes are arranged among the Q, Ψ=4 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3, l ₄=4,

The 357th step: check r ₂[v ₁], r ₂[v ₂], r ₂[v ₃], r ₂[v ₄], be respectively ∞, 0.0, ∞, so ∞ is l _λ=2, make u=v ₂,

The 358th step: establishing node u has Ω adjacent node, u=v ₂, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 359th step: establish cycle index g=1,

The 360th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 361st step: judge

This moment j=2, g=1, m _g=1, r _j[m _g]=∞, u=v ₂, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞1.1, and correct judgment then continues to carry out,

The 362nd step: order

Be r ₂[v ₁]=0.0+1.1=1.1,

The 363rd step: order

Be p ₂[v ₁]=v ₂,

The 364th step: make g=g+1, thus g=2,

The 365th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 366th step: judge This moment j=2, g=2, m _g=3, r _j[m _g]=∞, u=v ₂, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞3.0, and correct judgment then continues to carry out,

The 367th step: order

Be r ₂[v ₃]=0.0+3.0=3.0,

The 368th step: order

Be p ₂[v ₃]=v ₂,

The 369th step: make g=g+1, thus g=3,

The 370th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 371st step: u is moved to S set, u=v by the Q set ₂,

The 372nd step: judge whether current set Q is empty, at this moment Q={v ₁, v ₃, v ₄, 3 nodes are arranged among the Q, Ψ=3 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=3, l ₃=4,

The 373rd step: check r ₂[v ₁], r ₂[v ₃], r ₂[v ₄], being respectively 1.1,3.0, so ∞ is l _λ=1, make u=v ₁,

The 374th step: establishing node u has Ω adjacent node, u=v ₁, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 375th step: establish cycle index g=1,

The 376th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 377th step: judge

This moment j=2, g=1, m _g=2, r _j[m _g]=0.0, u=v ₁, r _j[u]=1.1, Therefore, the judgement formula is 0.0＞2.1, and next adjacent node is judged in misjudgment

The 378th step: make g=g+1, thus g=2,

The 379th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 380th step: judge

This moment j=2, g=2, m _g=4, r _j[m _g]=∞, u=v ₁, r _j[u]=1.1,

Therefore, the judgement formula is ∞＞14.8, and correct judgment then continues to carry out,

The 381st step: order

Be r ₂[v ₄]=0.0+13.7=13.7,

The 382nd step: order

Be p ₂[v ₄]=v ₁,

The 383rd step: make g=g+1, thus g=3,

The 384th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 385th step: u is moved to S set, u=v by the Q set ₁,

The 386th step: judge whether current set Q is empty, at this moment Q={v ₃, v ₄, 2 nodes are arranged among the Q, Ψ=2 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=3, l ₂=4,

The 387th step: check r ₂[v ₃], r ₂[v ₄], be respectively 3.0,14.8, so l _λ=3, make u=v ₃,

The 388th step: establishing node u has Ω adjacent node, u=v ₃, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 389th step: establish cycle index g=1,

The 390th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 391st step: judge

This moment j=2, g=1, m _g=2, r _j[m _g]=0.0, u=v ₃, r _j[u]=3.0,

Therefore, the judgement formula is 0.0＞6.0, and next adjacent node is judged in misjudgment

The 392nd step: make g=g+1, thus g=2,

The 393rd step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 394th step: judge

This moment j=2, g=2, m _g=4, r _j[m _g]=14.8, u=v ₃, r _j[u]=3.0,

Therefore, the judgement formula is 14.8＞4.0, and correct judgment then continues to carry out,

The 395th step: order Be r ₂[v ₄]=0.0+1.0=1.0,

The 396th step: order

Be p ₂[v ₄]=v ₃,

The 397th step: make g=g+1, thus g=3,

The 398th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 399th step: u is moved to S set, u=v by the Q set ₃,

The 400th step: judge whether current set Q is empty, at this moment Q={v ₄, 1 node is arranged among the Q, Ψ=1 according to node subscript series arrangement from small to large, obtains l to the node among the current Q ₁=4,

The 401st step: check r ₂[v ₄], be respectively 4.0, so l _λ=4, make u=v ₄,

The 402nd step: establishing node u has Ω adjacent node, u=v ₄, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 403rd step: establish cycle index g=1,

The 404th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 405th step: judge

This moment j=2, g=1, m _g=1, r _j[m _g]=1.1, u=v ₄, r _j[u]=4.0,

Therefore, the judgement formula is 1.1＞6.0, and next adjacent node is judged in misjudgment

The 406th step: make g=g+1, thus g=2,

The 407th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 408th step: judge

This moment j=2, g=2, m _g=3, r _j[m _g]=3.0, u=v ₄, r _j[u]=4.0, Therefore, the judgement formula is 3.0＞6.2, and next adjacent node is judged in misjudgment

The 409th step: make g=g+1, thus g=3,

The 410th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 411st step: u is moved to S set, u=v by the Q set ₄,

The 412nd step: judge whether current set Q is empty, and this moment, Q=φ then jumped out circulation,

The 413rd step: judge j≤τ, this moment j=3, τ=4, the judgement formula is 3≤4, correct judgment continues to carry out,

The 414th step: use single source impedance method of formation to calculate v ₃Impedance r to all nodes ₃[v _k] and with respect to v ₃Preposition node,

The 415th step: initialization p ₃[] array, r ₃[] array, S set, set Q,

The 416th step: p ₃[v ₁]=φ, p ₃[v ₂]=φ, p ₃[v ₃]=φ, p ₃[v ₄]=φ,

The 417th step: r ₃[v ₁]=∞, r ₃[v ₂]=∞, r ₃[v ₃]=0, r ₃[v ₄]=∞,

The 418th step: S=φ, Q={v ₁, v ₂, v ₃, v ₄,

The 419th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, v ₄, 4 nodes are arranged among the Q, Ψ=4 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3, l ₄=4,

The 420th step: check r ₃[v ₁], r ₃[v ₂], r ₃[v ₃], r ₃[v ₄], be respectively ∞, ∞, 0.0, so ∞ is l _λ=3, make u=v ₃,

The 421st step: establishing node u has Ω adjacent node, u=v ₃, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 422nd step: establish cycle index g=1,

The 423rd step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 424th step: judge

This moment j=3, g=1, m _g=2, r _j[m _g]=∞, u=v ₃, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞3.0, and correct judgment then continues to carry out,

The 425th step: order

Be r ₃[v ₂]=0.0+3.0=3.0,

The 426th step: order

Be p ₃[v ₂]=v ₃,

The 427th step: make g=g+1, thus g=2,

The 428th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 429th step: judge

This moment j=3, g=2, m _g=4, r _j[m _g]=∞, u=v ₃, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞1.0, and correct judgment then continues to carry out,

The 430th step: order

Be r ₃[v ₄]=0.0+1.0=1.0,

The 431st step: order

Be p ₃[v ₄]=v ₃,

The 432nd step: make g=g+1, thus g=3,

The 433rd step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 434th step: u is moved to S set, u=v by the Q set ₃,

The 435th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₄, 3 nodes are arranged among the Q, Ψ=3 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=4,

The 436th step: check r ₃[v ₁], r ₃[v ₂], r ₃[v ₄], be respectively ∞, 3.0,1.0, so l _λ=4, make u=v ₄,

The 437th step: establishing node u has Ω adjacent node, u=v ₄, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 438th step: establish cycle index g=1,

The 439th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 440th step: judge This moment j=3, g=1, m _g=1, r _j[m _g]=∞, u=v ₄, r _j[u]=1.0,

Therefore, the judgement formula is ∞＞3.0, and correct judgment then continues to carry out,

The 441st step: order

Be r ₃[v ₁]=0.0+2.0=2.0,

The 442nd step: order

Be p ₃[v ₁]=v ₄,

The 443rd step: make g=g+1, thus g=2,

The 444th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 445th step: judge

This moment j=3, g=2, m _g=3, r _j[m _g]=0.0, u=v ₄, r _j[u]=1.0,

Therefore, the judgement formula is 0.0＞3.2, and next adjacent node is judged in misjudgment

The 446th step: make g=g+1, thus g=3,

The 447th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 448th step: u is moved to S set, u=v by the Q set ₄,

The 449th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, 2 nodes are arranged among the Q, Ψ=2 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2,

The 450th step: check r ₃[v ₁], r ₃[v ₂], be respectively 3.0,3.0, so l _λ=1, make u=v ₁,

The 451st step: establishing node u has Ω adjacent node, u=v ₁, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=2, m ₂=4,

The 452nd step: establish cycle index g=1,

The 453rd step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 454th step: judge This moment j=3, g=1, m _g=2, r _j[m _g]=3.0, u=v ₁, r _j[u]=3.0,

Therefore, the judgement formula is 3.0＞4.0, and next adjacent node is judged in misjudgment

The 455th step: make g=g+1, thus g=2,

The 456th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 457th step: judge This moment j=3, g=2, m _g=4, r _j[m _g]=1.0, u=v ₁, r _j[u]=3.0,

Therefore, the judgement formula is 1.0＞16.7, and next adjacent node is judged in misjudgment

The 458th step: make g=g+1, thus g=3,

The 459th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 460th step: u is moved to S set, u=v by the Q set ₁,

The 461st step: judge whether current set Q is empty, at this moment Q={v ₂, 1 node is arranged among the Q, Ψ=1 according to node subscript series arrangement from small to large, obtains l to the node among the current Q ₁=2,

The 462nd step: check r ₃[v ₂], be respectively 3.0, so l _λ=2, make u=v ₂,

The 463rd step: establishing node u has Ω adjacent node, u=v ₂, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 464th step: establish cycle index g=1,

The 465th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 466th step: judge

This moment j=3, g=1, m _g=1, r _j[m _g]=3.0, u=v ₂, r _j[u]=3.0,

Therefore, the judgement formula is 3.0＞4.1, and next adjacent node is judged in misjudgment

The 467th step: make g=g+1, thus g=2,

The 468th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 469th step: judge

This moment j=3, g=2, m _g=3, r _j[m _g]=0.0, u=v ₂, r _j[u]=3.0,

Therefore, the judgement formula is 0.0＞6.0, and next adjacent node is judged in misjudgment

The 470th step: make g=g+1, thus g=3,

The 471st step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 472nd step: u is moved to S set, u=v by the Q set ₂,

The 473rd step: judge whether current set Q is empty, and this moment, Q=φ then jumped out circulation,

The 474th step: judge j≤τ, this moment j=4, τ=4, the judgement formula is 4≤4, correct judgment continues to carry out,

The 475th step: use single source impedance method of formation to calculate v ₄Impedance r to all nodes ₄[u _k] and with respect to v ₄Preposition node,

The 476th step: initialization p ₄[] array, r ₄[] array, S set, set Q,

The 477th step: p ₄[v ₁]=φ, p ₄[v ₂]=φ, p ₄[v ₃]=φ, p ₄[v ₄]=φ,

The 478th step: r ₄[v ₁]=∞, r ₄[v ₂]=∞, r ₄[v ₃]=∞, r ₄[v ₄]=0,

The 479th step: S=φ, Q={v ₁, v ₂, v ₃, v ₄,

The 480th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, v ₄, 4 nodes are arranged among the Q, Ψ=4 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3, l ₄=4,

The 481st step: check r ₄[v ₁], r ₄[v ₂], r ₄[v ₃], r ₄[v ₄], be respectively ∞, ∞, ∞, 0.0, so l _λ=4, make u=v ₄,

The 482nd step: establishing node u has Ω adjacent node, u=v ₄, then u has 2 adjacent nodes, Ω=2, to the adjacent node of node u according to subscript series arrangement from small to large, and will order corresponding to m ₁, m ₂, obtain m ₁=1, m ₂=3,

The 483rd step: establish cycle index g=1,

The 484th step: judge g≤Ω, this moment g=1, Ω=2, correct judgment then continues to carry out,

The 485th step: judge

This moment j=4, g=1, m _g=1, r _j[m _g]=∞, u=v ₄, r _j[u]=0.0,

Therefore, the judgement formula is ∞＞2.0, and correct judgment then continues to carry out,

The 486th step: order Be r ₄[v ₁]=0.0+2.0=2.0,

The 487th step: order

Be p ₄[v ₁]=v ₄,

The 488th step: make g=g+1, thus g=2,

The 489th step: judge g≤Ω, this moment g=2, Ω=2, correct judgment then continues to carry out,

The 490th step: judge

This moment j=4, g=2, m _g=3, r _j[m _g]=∞, u=v ₄, r _j[u]=0.0, Therefore, the judgement formula is ∞＞2.2, and correct judgment then continues to carry out,

The 491st step: order

Be r ₄[v ₃]=0.0+2.2=2.2,

The 492nd step: order

Be p ₄[v ₃]=v ₄,

The 493rd step: make g=g+1, thus g=3,

The 494th step: judge g≤Ω, this moment g=3, Ω=2, circulation is then jumped out in misjudgment,

The 495th step: u is moved to S set, u=v by the Q set ₄,

The 496th step: judge whether current set Q is empty, at this moment Q={v ₁, v ₂, v ₃, 3 nodes are arranged among the Q, Ψ=3 according to node subscript series arrangement from small to large, obtain l to the node among the current Q ₁=1, l ₂=2, l ₃=3,

The 497th step: check r ₄[v ₁], r ₄[v ₂], r ₄[v ₃], be respectively 2.0, ∞, 2.2, so l _λ=1, make u=v ₁,

The 558th step: finding source point is v ₂Node v ₃Preposition node p ₂[v ₃]=v ₂, node v then ₃With node v ₂Between the unidirectional highway section of unidirectional highway section impedance minimum be e ₃,

The 559th step: at the unidirectional highway section e of unidirectional highway section impedance minimum ₃Last distribute traffic amount, order accumulative total one-way road section traffic volume amount

$x_{{\mathrm{<figure-callout\; id="3"\; label="e"\; filenames="DEST\_PATH\_HSB00000294871800011.png,DEST\_PATH\_HSB00000294871800051.png"\; state="\{\{state\}\}">e</figure-callout>}}_3}={\mathrm{<figure-callout\; id="3"\; label="x\; e"\; filenames="DEST\_PATH\_HSB00000294871800011.png,DEST\_PATH\_HSB00000294871800051.png"\; state="\{\{state\}\}">x</figure-callout>}}_{e_{}}+50.00=150.00,$

The 560th step: record is to the unidirectional highway section e of highway section impedance minimum ₃On the volume of traffic constitute, the order accumulative total volume of traffic constitutes

$T_{e_3}\left(\mathrm{2,4}\right)=T_{e_3}\left(\mathrm{2,4}\right)+50.00=50.00,$

The 561st step: make pointer s=s '=2,

The 562nd step: judge s=o _n, this moment n=2, s=2, o _n=2, correct judgment is then jumped out circulation,

The 563rd step: make n=n+1, thus n=3,

The 564th step: judge n≤Γ, this moment n=3, Γ=2, circulation is then jumped out in misjudgment,

The 565th step: judge i≤∏, this moment i=3, ∏=2, the judgement formula is 3≤2, misjudgment, then whole process finishes, and obtains one-way road section traffic volume and forms T.

Claims (1)

1. the assay method that constitutes of an one-way road section traffic volume is characterized in that:

Step 1: the initialization transport need on average is split as ∏ five equilibrium: OD with the OD matrix of expressing transport need ₁Matrix, OD ₂Matrix ..., OD _∏Matrix, ∏ are the positive integer greater than 1, the OD matrix after each splits comprise Γ starting and terminal point to transport need, i.e. q _i(o ₁, d ₁), q _i(o ₂, d ₂) ..., q _i(o _n, d _n) ..., q _i(o _Γ, d _Γ), q wherein _i(o _n, d _n) expression OD _iStarting point is a node in the matrix

, terminal point is node

Transport need,

Step 2: with urban road network be set at resistive directed networks G (V, E), with the terminal in crossing in the urban road network and unidirectional highway section as described resistive directed networks G (V, node E) form set of node V, are designated as:

V={ υ ₁, υ ₂..., υ _τ, τ is the number of node, υ ₁, υ ₂, υ _τBe the node of directed networks, with the unidirectional highway section in the urban road network as described resistive directed networks G (the limit collection is designated as for V, directed edge E):

E={e ₁, e ₂..., e _σ..., e _η, e ₁, e ₂..., e _σ..., e _ηThe unidirectional highway section of the directed edge representative of expression directed networks, σ is the subscript in a certain unidirectional highway section, and η is the number in unidirectional highway section in the directed networks, and (V E) carries out initialization, obtains initialization directed networks G to directed networks G ₀(V, E),

Step 3: initialization impedance, the volume of traffic and the volume of traffic constitute, and make directed networks G ₀(V, E) in the accumulative total volume of traffic on all unidirectional highway sections

σ=1,2 ..., η,

Represent unidirectional highway section e _σOn the accumulative total volume of traffic; Make the volume of traffic on all unidirectional highway sections constitute

σ=1,2 ..., η,

Represent unidirectional highway section e _σGo up by node υ _oTo υ _dThe volume of traffic, the impedance in unidirectional highway section is:

$w_{e_{\mathrm{\&sigma;}}}=w_{e_{\mathrm{\&sigma;}}}^0(1+\mathrm{\&alpha;}{\left(\frac{x_{e_{\mathrm{\&sigma;}}}}{c_{e_{\mathrm{\&sigma;}}}}\right)}^{\mathrm{\&beta;}})$

Wherein

For when on the unidirectional highway section during without any vehicle, the time cost that vehicle passes through, it is that the geometrical length in known unidirectional highway section is divided by the known design speed of a motor vehicle; Be the one-way road section traffic volume amount, Be the maximum traffic capacity in unidirectional highway section, Be unidirectional highway section saturation degree; α and β are empirical parameter, α=0.15, and β=4.0,

Step 4: loop initialization pointer i=1,

Step 5: judge i≤∏, if execution in step 6 then; Otherwise, finishing, the volume of traffic that obtains on every highway section constitutes σ=1,2 ..., η, 1≤o≤τ, 1≤d≤τ,

Step 6: calculate current impedance matrix and be source point and with any node respectively with the impedance matrix generating algorithm, search shortest path for traffic in distributing and prepare with respect to the preposition node of another node arbitrarily,

Step 7: use the traffic allocation algorithm with transport need OD _iMatrix allocation arrives last time directed networks G _I-1(V, E) on, form current directed networks G _i(V E), and records to network G _i(V, E) volume of traffic on every the unidirectional highway section on constitutes,

Step 8: make i=i+1, return step 5,

Described impedance matrix generating algorithm is:

Step 6.1: loop initialization pointer j=1,

Step 6.2: if j≤τ, then execution in step 6.3, otherwise impedance matrix generates and to finish, and obtains the impedance between each node, and to obtain respectively with any node be source point and with respect to the preposition node of another node arbitrarily,

Step 6.3: adopt single source impedance generating algorithm to calculate arbitrary node υ _jArrive all node υ as source point _kImpedance r _j[υ _k], k=1,2 ... τ, and obtain preposition node with respect to arbitrary node of source point,

Step 6.4: make j=j+1, return step 6.2,

Step 6.3.1: initialization p _j[] array, r _j[] array, S set, set Q, p _j[υ _k] represent with node υ _jNode υ for source point _kPreposition node, k=1,2 ... τ, the order:

p _j[υ _k]＝φ

r _j[υ _k] represent with source point υ _jBe starting point, node υ _kBe the path impedance of terminal point, k=1,2 ... τ, order:

$r_j\left[{\mathrm{\&upsi;}}_k\right]=\left\{\begin{array}{cc}0& k=j\\ \&infin;& k\&NotEqual;j\end{array}\right.$

S set is used to deposit the node of having handled, initialization S, order

Set Q is used to deposit untreated node, and initialization Q makes Q=V,

Step 6.3.2: judge that whether the Q set is empty, if be empty, then obtains node υ _jArrive all node υ as source point _kImpedance r _j[υ _k], k=1,2 ..., τ, and obtain preposition node with respect to arbitrary node of source point; Otherwise, execution in step 6.3.3,

Step 6.3.3: establishing among the current set Q has ψ node, and ψ≤τ according to node subscript series arrangement from small to large, and corresponds respectively to l with tactic node subscript to the node among the current Q ₁, l ₂..., l _ψ, obtain

1≤l _ζ≤ τ,

Step 6.3.4: check

Therefrom find out minimum

And with node

As node u,

Step 6.3.5: determine source point υ respectively _jTo the impedance of each adjacent node of node u, and definite source point υ _jThe preposition node of adjacent node of node u,

Step 6.3.6: node u is moved to the S set by the Q set, execution in step 6.3.2,

Wherein, the described definite respectively source point υ of step 6.3.5 _jTo the impedance of each adjacent node of node u, and definite source point υ _jThe method of preposition node of adjacent node of node u be:

Step 6.3.5.1: establishing node u has Ω adjacent node, and Ω≤τ according to subscript series arrangement from small to large, and corresponds respectively to m with tactic node subscript to the adjacent node of node u ₁, m ₂..., m _Ω, obtain the adjacent node of tactic node u

1≤m _ρ≤ τ,

Step 6.3.5.2: establish cycle index g=1,

Step 6.3.5.3: if g≤Ω, execution in step 6.3.5.4 then, otherwise, current source point υ obtained _jTo the impedance of each adjacent node of node u, and source point υ _jThe preposition node of adjacent node of node u,

Step 6.3.5.4: if

Execution in step 6.3.5.5 then; Otherwise, skip to step

6.3.5.7, wherein, Expression by node u to adjacent node

The impedance of unidirectional highway section, if u is extremely

Between many unidirectional highway sections are arranged, then

Refer to the impedance on that unidirectional highway section of impedance minimum,

Step 6.3.5.5: make source point υ _jTo adjacent node

Impedance equal source point υ _jTo the impedance of u add u with

Impedance on the unidirectional highway section, even

Step 6.3.5.6: order

With respect to υ _jFor the preposition node of starting point is υ, even

Step 6.3.5.7: make g=g+1, return step 6.3.5.3,

Step 7 is described " uses the traffic allocation algorithm with transport need OD _iMatrix allocation to directed networks G last time (V, E) on, form current directed networks G (V, E), and directed networks G before the record (V, E) volume of traffic on every the unidirectional highway section on constitutes " method be:

Step 7.1: establish transport need OD _iComprise in the matrix Γ starting and terminal point to transport need, promptly

q _i(o ₁, d ₁), q _i(o ₂, d ₂) ..., q _i(o _n, d _n) ..., q _i(o _Γ, d _Γ), q wherein _i(o _n, d _n) expression OD _iStarting point is a node in the matrix

Terminal point is a node

Transport need,

Step 7.2: establish cycle index n=1,

Step 7.3: if n≤Γ, then execution in step 7.4; Otherwise, obtain current one-way road section traffic volume amount and road section traffic volume and form,

Step 7.4: with starting point-terminal point to allocation algorithm distribute the starting point that obtains by step 1

Terminal point

Between transport need q _i(o _n, d _n),

Step 7.5: make n=n+1, return step 7.3,

Wherein, the starting point-terminal point described in the step 7.4 to allocation algorithm be:

Step 7.4.1: establish pointer s=d _n, d _nBe the node subscript of the terminal point of point-to-point transmission transport need,

Step 7.4.2: if pointer s=o _n, then obtain transport need q _i(o _n, d _n) in starting point be Terminal point is

Shortest path on allocation result, otherwise, execution in step 7.4.3,

Step 7.4.3: find source point to be

Node υ _sPreposition node Node υ then _sWith source point be

Node υ _sPreposition node υ ' _sBetween the unidirectional highway section of unidirectional highway section impedance minimum be e _a,

Step 7.4.4: at the unidirectional highway section e of unidirectional highway section impedance minimum _aLast distribute traffic amount, order accumulative total one-way road section traffic volume amount $x_{e_a}=x_{e_a}+q_i(o_n,d_n),$

Step 7.4.5: record is to the unidirectional highway section e of highway section impedance minimum _aOn the volume of traffic constitute, the order accumulative total volume of traffic constitutes $T_{e_a}(o_n,d_n)=T_{e_a}(o_n,d_n)+q_i(o_n,d_n),$

Step 7.4.6: make pointer s=s ', return step 7.4.2 then.

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