数学建模常用Matlab/Lingo/c代码总结系列——旅行商TSP问题

Lingo代码：
MODEL: 

  SETS: 
  CITY / 1.. 6/: U; ! U( I) = sequence no. of city; 
  LINK( CITY, CITY): 
       DIST,  ! The distance matrix; 
          X;  ! X( I, J) = 1 if we use link I, J; 
 ENDSETS 
 DATA:   !Distance matrix, it need not be symmetric; 
  DIST =0 56 35 21 51 60 
 56 0 21 57 78 70 
 35 21 0 36 68 68 
 21 57 36 0 51 61 
 51 78 68 51 0 13 
 60 70 68 61 13 0; 
 ENDDATA 
 !The model:Ref. Desrochers & Laporte, OR Letters, 
  Feb. 91; 
  N = @SIZE( CITY); 
  MIN = @SUM( LINK: DIST * X); 
  @FOR( CITY( K): 
  !  It must be entered; 
   @SUM( CITY( I)| I #NE# K: X( I, K)) = 1; 
  !  It must be departed; 
   @SUM( CITY( J)| J #NE# K: X( K, J)) = 1; 
  ! Weak form of the subtour breaking constraints; 
  ! These are not very powerful for large problems; 
   @FOR( CITY( J)| J #GT# 1 #AND# J #NE# K: 
       U( J) >= U( K) + X ( K, J) - 
       ( N - 2) * ( 1 - X( K, J)) + 
       ( N - 3) * X( J, K))); 
  ! Make the X's 0/1; 
  @FOR( LINK: @BIN( X)); 
  ! For the first and last stop we know...; 
  @FOR( CITY( K)| K #GT# 1: 
   U( K) <= N - 1 - ( N - 2) * X( 1, K); 
   U( K) >= 1  + ( N - 2) * X( K, 1)); 
END 

matlab代码：function main clc,clear global a % a=zeros(6); % a(1,2)=56;a(1,3)=35;a(1,4)=21;a(1,5)=51;a(1,6)=60; % a(2,3)=21;a(2,4)=57;a(2,5)=78;a(2,6)=70; % a(3,4)=36;a(3,5)=68;a(3,6)=68; a(4,5)=51;a(4,6)=61; % a(5,6)=13; a=a+a'; load costa=Muti_Cost;%边权矩阵L=size(a,1); c1=1:53; %初始圈[circle,long]=modifycircle(c1,L); c2=[1 53 2:52];%改变初始圈，该算法的最后一个顶点不动 [circle2,long2]=modifycircle(c2,L); if long2<long    long=long2;    circle=circle2; end circle,long %******************************************* %修改圈的子函数 %******************************************* function [circle,long]=modifycircle(c1,L); global a flag=1; while flag>0       flag=0;    for m=1:L-3       for n=m+2:L-1         if a(c1(m),c1(n))+a(c1(m+1),c1(n+1))<...            a(c1(m),c1(m+1))+a(c1(n),c1(n+1))            flag=1;            c1(m+1:n)=c1(n:-1:m+1);         end      end   end end long=a(c1(1),c1(L)); for i=1:L-1    long=long+a(c1(i),c1(i+1)); end circle=c1;