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

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题目大意：
有n只奶牛，有n个连续的实力，如果u的实力大于v的实力，就能打赢它，
然后给定m种关系，求最后能确定其排名的奶牛个数。
分析：
传递闭包可以快速求解。用floyd，求出每个点能到达的点。如果有一个点，排名在它之前的和排名在它之后的点之和为n-1，那么它的排名就是确定的。

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#include<iostream>#include<cstdio>using namespace std;int n,m,ans;bool f[101][101];int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=m;i++)    {        int x,y;        scanf("%d%d",&x,&y);        f[x][y]=1;    }    for(int k=1;k<=n;k++)        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)                if(f[i][k]&&f[k][j])                    f[i][j]=1;    for(int i=1;i<=n;i++)    {        int res=n-1;        for(int j=1;j<=n;j++)            if(f[i][j]||f[j][i])                res--;        if(res==0)            ans++;    }    printf("%d\n",ans);    return 0;}