Rank HDU

there are N ACMers in HDU team.
ZJPCPC Sunny Cup 2007 is coming, and lcy want to select some excellent ACMers to attend the contest. There have been M matches since the last few days(No two ACMers will meet each other at two matches, means between two ACMers there will be at most one match). lcy also asks"Who is the winner between A and B?" But sometimes you can't answer lcy's query, for example, there are 3 people, named A, B, C.and 1 match was held between A and B, in the match A is the winner, then if lcy asks "Who is the winner between A and B", of course you can answer "A", but if lcy ask "Who is the winner between A and C", you can't tell him the answer.
As lcy's assistant, you want to know how many queries at most you can't tell lcy(ask A B, and ask B A is the same; and lcy won't ask the same question twice).

InputThe input contains multiple test cases.
The first line has one integer,represent the number of test cases.
Each case first contains two integers N and M(N , M <= 500), N is the number of ACMers in HDU team, and M is the number of matchs have been held.The following M lines, each line means a match and it contains two integers A and B, means A wins the match between A and B.And we define that if A wins B, and B wins C, then A wins C.OutputFor each test case, output a integer which represent the max possible number of queries that you can't tell lcy.Sample Input

3
3 3
1 2
1 3
2 3
3 2
1 2
2 3
4 2
1 2
3 4

Sample Output

0
0
4


        
 

Hint

in the case3, if lcy ask (1 3 or 3 1) (1 4 or 4 1) (2 3 or 3 2) (2 4 or 4 2), then you can't tell him who is the winner.

题意：有一场比赛，对于n个选手给出m场比赛的结果，结果有传递性，问有多少对选手不能判断他们之间的胜负关系。

思路 :floyd求解传递闭包

AC代码：

#include<bits/stdc++.h>

using namespace std;
#define maxn 766
int e[maxn][maxn];
int  main(){
    int _;
    cin>>_;
    while(_--){
        int n,m;
        scanf("%d%d",&n,&m);
        int x,y;
        while(m--){
            //cin>>x>>y;
            scanf("%d%d",&x,&y);
            e[x][y]=1;
        }
        for(int k=1;k<=n;k++)
            for(int i=1;i<=n;i++){
                if(!e[i][k])
                    continue; 
                for(int j=1;j<=n;j++)
                    if(e[i][k]&&e[k][j])
                        e[i][j]=1;
            }
                
                        
        int ans=0;
        for(int i=1;i<=n;i++)
            for(int j=i+1;j<=n;j++)
                if(!e[i][j]&&!e[j][i])
                    ans++;
        printf("%d
",ans);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                e[i][j]=0;
    }
}