POJ 3660 - Cow Contest(传递闭包+Floyd)

Cow Contest
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 10479 Accepted: 5853
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：
给出n的牛和m对关系（A牛可以战胜B牛），问最后有多少个牛的rank可以确定。

解题思路：
传递闭包+floyd，如果当前的牛和n-1头牛的关系都存在，那么这头牛的rank就知道了。
并且如果a和b有关系，b和c有关系，那么a和c也有关系。

AC代码：

#include<stdio.h>#include<string.h>const int maxn = 1e2+5;bool mp[maxn][maxn];int main(){    int n,m;    scanf("%d%d",&n,&m);    memset(mp,0,sizeof(mp));    while(m--)    {        int a,b;        scanf("%d%d",&a,&b);        mp[a][b] = 1;    }    for(int k = 1;k <= n;k++)        for(int i = 1;i <= n;i++)            for(int j = 1;j <= n;j++)                if(mp[i][k] && mp[k][j])    mp[i][j] = 1;    int ans = 0;    int j;    for(int i = 1;i <= n;i++)    {        for(j = 1;j <= n;j++)        {            if(i == j)  continue;            if(!mp[i][j] && !mp[j][i])  break;        }        if(j > n)   ans++;    }    printf("%d",ans);    return 0;}