中国邮递员问题matlab.doc
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%中国邮递员问题:
%step1;
%求出奇点之间的距离;
%求各个点之间的最短距离;
%floyd算法;
clear all;
clc;
A=zeros(9);
A(1,2)=3; A(1,4)=1;
A(2,4)=7; A(2,5)=4;A(2,6)=9;A(2,3)=2;
A(3,6)=2
A(4,7)=2; A(4,8)=3;A(4,5)=5;
A(5,6)=8;
A(6,9)=1;A(6,8)=6;
A(7,8)=2;
A(8,9)=2;
c=A+A;
c(find(c==0))=inf;
m=length(c);
Path=zeros(m);
for k=1:m
for i=1:m
for j=1:m
if c(i,j)c(i,k)+c(k,j)
c(i,j)=c(i,k)+c(k,j);
Path(i,j)=k;
end
end
end
end
c, Path
h1=c(2,4);
h2=c(2,6);
h3=c(2,5);
h4=c(4,6);
h5=c(4,5);
h6=c(6,5);
h=[h1,h2,h3,h4,h5,h6]
%step2;
%找出以奇点为顶点的完全图的最优匹配;
%算法函数Hung_Al.m
function [Matching,Cost] = Hung_Al(Matrix)
Matching = zeros(size(Matrix));
% 找出每行和每列相邻的点数
num_y = sum(~isinf(Matrix),1);
num_x = sum(~isinf(Matrix),2);
% 找出每行和每列的孤立点数
x_con = find(num_x~=0);
y_con = find(num_y~=0);
%将矩阵压缩、重组
P_size = max(length(x_con),length(y_con));
P_cond = zeros(P_size);
P_cond(1:length(x_con),1:length(y_con)) = Matrix(x_con,y_con);
if isempty(P_cond)
Cost = 0;
return
end
% 确保存在完美匹配,计算矩阵边集
Edge = P_cond;
Edge(P_cond~=Inf) = 0;
cnum = min_line_cover(Edge);
Pmax = max(max(P_cond(P_cond~=Inf)));
P_size = length(P_cond)+cnum;
P_cond = ones(P_size)*Pmax;
P_cond(1:length(x_con),1:length(y_con)) = Matrix(x_con,y_con);
%主函数程序,此处将每个步骤用switch命令进行控制调用步骤函数
exit_flag = 1;
stepnum = 1;
while exit_flag
switch stepnum
case 1
[P_cond,stepnum] = step1(P_cond);
case 2
[r_cov,c_cov,M,stepnum] = step2(P_cond);
case 3
[c_cov,stepnum] = step3(M,P_size);
case 4
[M,r_cov,c_cov,Z_r,Z_c,stepnum] = step4(P_cond,r_cov,c_cov,M);
case 5
[M,r_cov,c_cov,stepnum] = step5(M,Z_r,Z_c,r_cov,c_cov);
case 6
[P_cond,stepnum] = step6(P_cond,r_cov,c_cov);
case 7
exit_flag = 0;
end
end
Matching(x_con,y_con) = M(1:length(x_con),1:length(y_con));
Cost = sum(sum(Matrix(Matching==1)));
%下面是6个步骤函数step1~step6
%步骤1:找
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