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

Sample Output

Source

USACO 2008 January Silver

思路：这道题用的是拓扑建图，用佛洛依德建立每个点与其他点的关系。如果能确定这个点与其他所有点的关系，那它就是可以明确排名的点，否则就无法建立关系——无法确定排名（自己画图）。

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define ll long long

const int inf = 0x3f3f3f3f;
int n, m, dis[100+8][100+8];

int main()
{
    int a, b;
    memset(dis, 0, sizeof(dis));
    scanf("%d%d", &n, &m);
    for(int i = 0; i<m; i++)
    {
        scanf("%d%d", &a, &b);
        dis[a][b] = 1;
        dis[b][a] = -1;
    }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            for(int k = 1; k <= n; k++)
                if(dis[j][i] == dis[i][k] && dis[j][i])
                {
                    dis[j][k] = dis[i][k];
                }
    int ans = 0, sum;
    for(int i = 1; i <= n; i++)
    {
        sum = 0;
        for(int j = 1; j <= n; j++)
            if(dis[i][j] != 0)
                sum++;
        if(sum == n-1)ans++;
    }
    printf("%d\n", ans);
    return 0;
}

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