flyod 求传递闭包 （离散数学及应用p638 26）

 Write a program to implement  Computing the Transitive Closure，and testing you program with p638 26 in the text book.

26. Use your Algorithm to find the transitive closures of these relations on {a, b, c, d, e}.

a) {(a, c), (b, d), (c, a), (d, b), (e, d)}

b) {(b, c), (b, e), (c, e), (d, a), (e, b), (e, c)}

c) {(a, b), (a, c), (a, e), (b, a), (b, c), (c,a), (c,b), (d,a),(e, d)}

d) {(a, e), (b, a), (b, d), (c,d), (d,a), (d, c), (e,a), (e,b),(e, c), (e, e)}

 

借助flyod算法，用第三个顶点更新R[i][j],

当R[i][k]&R[k][j]都为1时R[i][j]增加一条边

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
const int maxn = 100 + 5;
int n,T;
bool R[maxn][maxn],flag;
void Init()
{
    memset(R,0,sizeof(R));
    flag=n=0;
    cout<<"Please enter how many elements there are in the relationship"<<endl;
    cin>>n;
    cout<<"Please enter the elements of the relationship"<<endl;
    for(int i=1;i<=n;i++)
    {
         char a,b;
         cin>>a>>b;
         R[a-'a'+1][b-'a'+1] = true;
    }
}
int main()
{
    cout<<"Please enter the number of data sets"<<endl;
    cin>>T;
    while(T--)
    {
        Init();
        for(int k=1;k<=5;k++)
            for(int i=1;i<=5;i++)
                for(int j=1;j<=5;j++)        
                    R[i][j]|=R[i][k]&R[k][j];//flyod
        printf("the transitive closures of R is{");
        for(int i=1;i<=5;i++)
            for(int j=1;j<=5;j++)
            {
                if(R[i][j]) printf("(%c,%c)",i-1+'a',j-1+'a');
            }
        printf("}\n");
    }
    return 0;
}
/*
4
5
a c
b d
c a
d b
e d
6
b c
b e 
c e
d a 
e b 
e c
9
a b
a c
a e
b a
b c
c a
c b
d a
e d
10
a e
b a
b d
c d
d a
d c
e a
e b
e c
e e
*/