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;
