Cow Contest

Time Limit: 1000MS Memory Limit: 65536K

Description

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

思路：

Floyd将所有连通路都求出来，然后进行每条边的遍历，如果和其他边都有路，那么就可以肯定这个的等级了。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 110;
bool map[maxn][maxn] = {false};
int main() {
    ios::sync_with_stdio(false);
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        map[a][b] = true;
    }
    for (int k = 1; k <= n; k++) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if (map[i][k] && map[k][j]) map[i][j] = true;
            }
        }
    }
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        int ans = 1;
        bool flag = true;
        for (int j = 1; j <= n; j++) {
            if (map[i][j] || map[j][i]) ans++;
        }
        if (ans == n) sum++;
    }
    printf("%d\n", sum);
    return 0;
}