Cow Contest [最短路][floyd]

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

• Line 1: Two space-separated integers: N and M
• Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

• Line 1: A single integer representing the number of cows whose ranks can be determined
  　

Sample Input

5 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2

题解

若点A与除A外的所有点关系确定(联通)，则这点的rank是确定的

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
#define MAX_N 102
#define INF 0x3f3f3f3f
using namespace std;
int d[MAX_N][MAX_N];
int N,M;

void solve(){
    for(int i=1;i<=N;i++)
        for(int j=1;j<=N;j++)
            for(int k=1;k<=N;k++)
                if(d[j][k]>=d[j][i]+d[i][k])
                    d[j][k]=d[j][i]+d[i][k];
    int cum=0;
    for(int i=1;i<=N;i++){
        int res=0;
        for(int j=1;j<=N;j++){
            if(d[i][j]!=INF||d[j][i]!=INF) res++;
            else break;
        }
        if(res==N) cum++;
    }
    printf("%d\n",cum);
}

int main()
{
    while(~scanf("%d%d",&N,&M)){
        memset(d,0x3f,sizeof(d));
        for(int i=1;i<=N;i++) d[i][i]=0;

        for(int i=0,x,y;i<M;i++){
            scanf("%d%d",&x,&y);
            d[x][y]=1;
        }
        solve();
    }
    return 0;
}