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 54 34 23 21 22 5

Sample Output

2

Source

USACO 2008 January Silver

题意：

      有n头牛比赛，m次比赛结果，问你一共有多少头牛的排名可以被确定，其中如果i战胜k，k战胜j，则也可以说i战胜j，即可以传递胜负。

分析：

这是一道经典的Floyd算法求传递闭包问题。所谓传递性，可以这样理解：对于一个节点i，如果i能到k，k能到j，那么i就能到j。求传递闭包，就是把图中所有满足这样传递性的节点都求出来，计算完成后，我们也就知道了任意两个节点之间是否相连。

   本题中，如果一头牛被x头牛打败，打败y头牛，且x+y=n-1，则这头牛的排名被确定了。我们首先确定一下任何两头牛的胜负关，再遍历所有牛判断一下是否满足x+y=n-1，将满足这个条件的牛数目加起来就是所求解。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int inf=0x3f3f3f3f;const int Max_n=110;int n,m;int d[Max_n][Max_n];int main(){    scanf("%d%d",&n,&m);    memset(d,0,sizeof(d));    int a,b;    for(int i=1;i<=m;i++){        scanf("%d%d",&a,&b);        d[a][b]=1;    }    for(int k=1;k<=n;k++) //求出任意两头牛的胜负关系        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++){                d[i][j]=d[i][j]||(d[i][k]&&d[k][j]);            }    int ans=0;    for(int i=1;i<=n;i++){        int cnt=0;        for(int j=1;j<=n;j++)            if(d[i][j]||d[j][i])cnt++;        if(cnt==n-1)ans++;    }    printf("%d\n",ans);    return 0;}