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
题意：有些奶牛的遗传因子比较好，导致它的能力比较强，现在有n头奶牛，告诉你m对奶牛的能力比较结果，输入a和b，表示a的能力比b的强，那么最终可以确定名次的奶牛个数是多少。（又是奶牛，真是偏爱啊~~）
分析一下，我们先进行一下最短路径的计算（用Floyd算法最合适），那么只要G[i][j]不为INF，就代表i的能力比j的强，那么当不为INF都情况=n-1时，表明这个奶牛和其它奶牛的关系确定了，那么它的名次也就出来了，统计这样的奶牛个数就OK了。

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

const int N=110;
const int INF=0x3f3f3f3f;

int G[N][N], n;

void Init()
{
    int i, j;

    for (i = 1; i <= n; i++)
    {
        for (j = 1; j <= n; j++)
            G[i][j] = INF;
        G[i][i] = 0;
    }
}

void Dist() ///用Floyd算法可以将每两个奶牛的关系表示出来
{
    int i, j, k;

    for (k = 1; k <= n; k++)
    {
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
                G[i][j] = min(G[i][j], G[i][k]+G[k][j]);
        }
    }
}

int main ()
{
    int m, a, b, c[N], ans, i, j;

    while (scanf("%d%d", &n, &m) != EOF)
    {
        Init();
        memset(c, 0, sizeof(c)); ///c数组保存确定关系的次数
        ans = 0;

        while (m--)
        {
            scanf("%d%d", &a, &b);
            G[a][b] = 1;
        }

        Dist();

        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
            {
                if (G[i][j] != INF && i != j) ///要是相等了就没有意义了
                {
                    c[i]++;
                    c[j]++;
                }
            }
        }

        for (i = 1; i <= n; i++)
        {
            if (c[i] == n-1)
                ans++;
        }

        printf("%d\n", ans);
    }

    return 0;
}