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 cost
a=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;