poj 3660 Cow Contest（warshall算法）

poj 3660 Cow Contest

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

题目大意：一群牛比赛，每场两只牛对打，并分出胜负，现在问你能确定几只牛的战斗力排名。

解题思路：warshall算法。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;

typedef long long ll;
const int N = 105;
int G[N][N];
int n, m, t;
void floyd() {
    for (int k = 1; k <= n; k++) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                G[i][j] = G[i][j] || (G[i][k] & G[k][j]);   
            }   
        }   
    }
}
int main() {
    while (scanf("%d %d", &n, &m) != EOF) {
        memset(G, 0, sizeof(G));
        int a, b;
        for (int i = 0; i < m; i++) {
            scanf("%d%d", &a, &b);      
            G[a][b] = 1;
        }   
        floyd();
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            int cnt = 0;    
            for (int j = 1; j <= n; j++) {
                if (G[i][j]) cnt++;
                if (G[i][j]) cnt++;
            } 
            if (cnt == n - 1) ans++;
        }
        printf("%d\n", ans);
    }   
    return 0;
}