Cow Contest POJ

Cow Contest

POJ - 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 cowA has a greater skill level than cowB (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠B), then cow A will always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM (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

题意：

题目给出了m对关系，求有多少个点的顺序关系是确定的。

思路：只要是这个点个其他所有点的关系是确定点，那么这个点的次序就是确定了，进一步简化，也就是看这个点是否和其他所有点有关系（边）不用管边的方向，因为不管什么方向，只要右边就可以确定一种关系，所以这个题就可以用floyd求传递闭包来做

code：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int n,m;int G[120][120];int main(){    int i,j;    cin >> n >> m;    memset(G,0,sizeof(G));    for(i = 1; i <= m; i++){        int u,v;        cin >> u >> v;        G[u][v] = 1;    }    int k;    //floyd求传递闭包的核心算法    for(k = 1; k <= n; k++){        for(i = 1; i <= n; i++){            for(j = 1; j <= n; j++){                G[i][j] = G[i][j] || (G[i][k] && G[k][j]);            }        }    }    //把每个点都判断一下，看是否可以确定关系    int ans = 0;    for(i = 1; i <= n; i++){        for(j = 1; j <= n; j++){            if(i == j)continue;            if(!G[i][j] && !G[j][i])                break;        }        if(j > n)            ans++;    }    cout << ans << endl;    return 0;}