POJ 3360 H-Cow Contest

http://poj.org/problem?id=3660

 

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：$floyd$ 算法 如果赢过的人加上败给的人的和是 $N - 1$ 就是可以确定位置的人

代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
using namespace std;

int N, M;
int mp[110][110];

void floyd() {
    for(int k = 1; k <= N; k ++) {
        for(int i = 1; i <= N; i ++) {
            if(mp[i][k])
            for(int j = 1; j <= N; j ++) {
                if(mp[k][j] == 1 && mp[i][k] == 1) {
                    mp[i][j] = 1;
                    mp[j][i] = -1;
                }
                else if(mp[k][j] == -1 && mp[i][k] == -1) {
                    mp[i][j] = -1;
                    mp[j][i] = 1;
                }
                else continue;
            }
        }
    }
}

int main() {
    memset(mp, 0, sizeof(mp));
    scanf("%d%d", &N, &M);
    for(int i = 1; i <= M; i ++) {
        int a, b;
        scanf("%d%d", &a, &b);
        mp[a][b] = 1;
        mp[b][a] = -1;
    }

    floyd();
    int ans = 0;
    for(int i = 1; i <= N; i ++) {
        int sum = 0;
        for(int j = 1; j <= N; j ++)
            if(mp[i][j])
                sum ++;

        if(sum == N - 1)
            ans ++;
    }
    printf("%d\n", ans);
    return 0;
}