例 9 某公司在六个城市 c1,c2, …c6 中有分公司，从 i
ci 到 cj的直接航程票价记在下述矩阵的 (I,j) 位置上。（∞表示无直接航路），请帮助该公司设计一张城市c1 到其它城市间的票价最便宜的路线图。
clc,clear
a=zeros(6);
a(1,2)=50;a(1,4)=40;a(1,5)=25;a(1,6)=10;
a(2,3)=15;a(2,4)=20;a(2,6)=25;
a(3,4)=10;a(3,5)=20;
a(4,5)=10;a(4,6)=25;
a(5,6)=55;
a=a+a';
a(find(a==0))=inf;
pb(1:length(a))=0;pb(1)=1;index1=1;index2=ones(1,length(a));
d(1:length(a))=inf;d(1)=0;temp=1;
while sum(pb)<length(a)
tb=find(pb==0);
d(tb)=min(d(tb),d(temp)+a(temp,tb));
tmpb=find(d(tb)==min(d(tb)));
temp=tb(tmpb(1));
pb(temp)=1;
index1=[index1,temp];
temp2=find(d(index1)==d(temp)-a(temp,index1));
index2(temp)=index1(temp2(1));
end
d, index1, index2
编写 LINGO 程序如下：
model:
sets:
cities/A,B1,B2,C1,C2,C3,D/;
roads(cities,cities)/A B1,A B2,B1 C1,B1C2,B1 C3,B2 C1,
B2 C2,B2 C3,C1 D,C2 D,C3 D/:w,x;
endsets
data:
w=2 4 3 3 1 2 3 1 1 3 4;
enddata
n=@size(cities); !城市的个数;
min=@sum(roads:w*x);
@for(cities(i)|i #ne#1 #and# i #ne#n:
@sum(roads(i,j):x(i,j))=@sum(roads(j,i):x(j,i)));
@sum(roads(i,j)|i #eq#1:x(i,j))=1;
@sum(roads(i,j)|j #eq#n:x(i,j))=1;
end
model:
sets:
cities/1..11/;
roads(cities,cities):w,x;
endsets
data:
w=0;
enddata
calc:
w(1,2)=2;w(1,3)=8;w(1,4)=1;
w(2,3)=6;w(2,5)=1;
w(3,4)=7;w(3,5)=5;w(3,6)=1;w(3,7)=2;
w(4,7)=9;
w(5,6)=3;w(5,8)=2;w(5,9)=9;
w(6,7)=4;w(6,9)=6;
w(7,9)=3;w(7,10)=1;
w(8,9)=7;w(8,11)=9;
w(9,10)=1;w(9,11)=2;w(10,11)=4;
@for(roads(i,j):w(i,j)=w(i,j)+w(j,i));
@for(roads(i,j):w(i,j)=@if(w(i,j) #eq# 0,1000,w(i,j)));
endcalc
n=@size(cities); !城市的个数;
min=@sum(roads:w*x);
@for(cities(i)|i #ne#1 #and# i #ne#
n:@sum(cities(j):x(i,j))=@sum(cities(j):x(j,i)));
@sum(cities(j):x(1,j))=1;
@sum(cities(j):x(j,1))=0; !不能回到顶点1;
@sum(cities(j):x(j,n))=1;
@for(roads:@bin(x));
end
例12 用Floyd算法求解例9。
矩阵path用来存放每对顶点之间最短路径上所经过的顶点的序号。Floyd算法的
Matlab程序如下：
clear;clc;
n=6; a=zeros(n);
a(1,2)=50;a(1,4)=40;a(1,5)=25;a(1,6)=10;
a(2,3)=15;a(2,4)=20;a(2,6)=25;a(3,4)=10;a(3,5)=20;
a(4,5)=10;a(4,6)=25; a(5,6)=55;
a=a+a'; M=max(max(a))*n^2; %M为充分大的正实数
a=a+((a==0)-eye(n))*M;
path=zeros(n);
for k=1:n
for i=1:n
for j=1:n
if a(i,j)>a(i,k)+a(k,j)
a(i,j)=a(i,k)+a(k,j);
path(i,j)=k;
end
end
end
end
a, path
我们使用LINGO9.0编写的FLOYD算法如下：
model:
sets: nodes/c1..c6/;
link(nodes,nodes):w,path; !path标志最短路径上走过的顶点;
endsets
data:
path=0;
w=0;
@text(mydata1.txt)=@writefor(nodes(i):@writefor(nodes(j):
@format(w(i,j),' 10.0f')),@newline(1));
@text(mydata1.txt)=@write(@newline(1));
@text(mydata1.txt)=@writefor(nodes(i):@writefor(nodes(j):
@format(path(i,j),' 10.0f')),@newline(1));
enddata
calc:
w(1,2)=50;w(1,4)=40;w(1,5)=25;w(1,6)=10;
w(2,3)=15;w(2,4)=20;w(2,6)=25;
w(3,4)=10;w(3,5)=20;
w(4,5)=10;w(4,6)=25;w(5,6)=55;
@for(link(i,j):w(i,j)=w(i,j)+w(j,i));
@for(link(i,j) |i#ne#j:w(i,j)=@if(w(i,j)#eq#0,10000,w(i,j)));
@for(nodes(k):@for(nodes(i):@for(nodes(j):
tm=@smin(w(i,j),w(i,k)+w(k,j));
path(i,j)=@if(w(i,j)#gt#tm,k,path(i,j));w(i,j)=tm)));
endcalc
end