数学建模--运输问题 下载本文

内容发布更新时间 : 2024/12/23 6:46:14星期一 下面是文章的全部内容请认真阅读。

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附录3:

model: sets:

v / 1.. 10/: u,D;

link( v, v):C,y, x; endsets

min = @sum( link: C*x);

@sum(v( I)| I #ne# 1: x(I, 1))=1; @sum(v( I)| I #ne# 1: x(1,I))=1; @FOR( v(K):

@sum( v( I)| I #ne# K: x( I, K))<1; @sum( v( J)| J #ne# K: x(K,J))<1); @FOR( v(K):

@bin(@sum( v( I)| I #ne# K: x( I, K)));

@bin(@sum( v( J)| J #ne# K: x(K,J)))); @sum(v(I)| I #ne#6 :x(I,6))=1; !@sum(v(I)| I #ne#10 :x(I,10))=1; !@sum(v(I)| I #ne#4 :x(I,4))=1; !@sum(v(I)| I #ne#2:x(I,2))=1; !@sum(v(I)| I #ne#3:x(I,3))=1; !@sum(v(I)| I #ne#5:x(I,5))=1; !@sum(v(I)| I #ne#8:x(I,8))=1; !@sum(v(I)| I #ne#7:x(I,7))=1; !@sum(link(I,J)|I #ne# J:x(I,J))>4 @sum(link(I,J)|I #ne# J:x(I,J)*D(J))>36 @sum(link(I,J)|I #ne# J:x(I,J)*D(J))<50; @for(v(I):

@sum(v(J): x(I,J))-@sum(v(J): x(J,I))=0); @for(v(I)|I #gt# 1:

@for( v( J)| J#gt#1 #and# I #ne# J: u(I)-u(J)+10*x(I,J)<=9)); @for( link: @bin( x)); @for( link: @gin( x)); data:

D=8,13,6,9,7,15,10,5,12,9;

C=0 50 999 40 25 999 30 999 50 999

50 0 30 999 35 50 999 60 999 999 999 30 0 15 999 30 50 25 999 60 40 999 15 0 45 30 55 20 40 65 25 15 999 45 0 60 10 30 999 55 999 50 30 30 60 0 25 55 35 999 30 999 50 999 10 25 0 30 45 60 999 60 25 20 30 55 30 0 10 999 20 999 999 40 999 15 25 45 0 20 35 20 10 45 20 999 60 999 30 0

; enddata end

附录4:

clear; clc;

M=10000;%不能直接到达是将距离赋值给M a(1,:)=[0,50,M,40,25,M,30,M,50,M]; a(2,:)=[50,0,30,M,35,50,M,60,M,M]; a(3,:)=[M,30,0,15,M,30,50,25,M,60]; a(4,:)=[40,M,15,0,45,30,55,20,40,65]; a(5,:)=[25,15,M,45,0,60,10,30,M,55]; a(6,:)=[M,50,30,30,60,0,25,55,35,M]; a(7,:)=[30,M,50,M,10,25,0,30,45,60]; a(8,:)=[M,60,25,20,30,55,30,0,10,M]; a(9,:)=[20,M,M,40,M,15,25,45,0,20];

a(10,:)=[35,20,10,45,20,M,60,M,30,0];%建立a矩阵

path=zeros(length(a)); %建立一个与矩阵a同大小的全零矩阵for i=1:10 for j=1:10

path(i,j)=j; %用path矩阵记录走过的点 end end

for k=1:10 for i=1:10 for j=1:10

if a(i,j)>a(i,k)+a(k,j) a(i,j)=a(i,k)+a(k,j);

path(i,j)=path(i,k); % floyd算法 end end end end a, path

i1=input('请输入起点'); i2=input('请输入终点'); disp(i1); while i1~=i2 i1=path(i1,i2); disp(i1); end; a =

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30 45 25 20 30 25 30 0 10 30 20 40 30 40 35 15 25 45 0 20 35 20 10 25 20 40 30 35 30 0 path =

1 5 4 4 5 7 7 5 9 9 1 2 3 3 5 6 5 3 3 3 4 2 3 4 8 6 7 8 8 8 1 3 3 4 5 6 8 8 8 8 1 2 2 4 5 7 7 8 8 10 7 2 3 4 7 6 7 4 9 9 1 5 3 8 5 6 7 8 8 10 9 5 3 4 5 9 7 8 9 9 1 10 10 4 7 6 7 8 9 10 1 2 3 3 5 3 5 3 9 10 请输入起点1 请输入终点2 1 5 2