Matlab求传递闭包t(R)

a=size(R)

B=zeros(a)

flag=0

while

flag==0

for

i=

1:

a

for

j=

1:

a

for

k=1:a

B(

i

,

j

)

=

max(min(

R(

i

,

k)

,

R(

k,

j)

)

,

B(

i

,

j

)

)

%R与昌侍R内积，先取小再取誉迅洞大

end

end

end

if

B==R

flag=1

else

R=B%循环计算R传递闭包庆枯

end

end

程序运行通过，很好用

#include<iostream>#include<memory>using namespace stdint main() { int r[101][101]int a=0,b=0,c=0,n=0cout<<"请输入关系矩阵的阶数"<<带郑endlcin>>nfor (a=1a<=na++) { cout<<"请输入悔指矩阵的第"<<a<<"行蠢前颂,中间一空格隔开."<<endlfor (b=1b<=nb++) cin>>r[a][b]} cout<<endl<<"原矩阵为"<<endlfor (a=1a<=na++) { for (b=1b<=nb++) cout<<r[a][b]<<" "cout<<endl} //以上输入矩阵完成 bool _finish=0for() { _finish=0for(a=1a<=na++) for(b=1b<=nb++) for(c=1c<=nc++) if ((r[a][b]==1)&&(r[b][c]==1)&&r[a][c]==0) { r[a][c]=1_finish=1} if (_finish ==0) break} cout<<endl<<"t(r)的关系矩阵为"<<endlfor(a=1a<=na++) { for(b=1b<=nb++) cout<<r[a][b]<<" "cout<<endl} return 0}

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