Cow Contest POJ - 3660（floyed求传递闭包）

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
思路：很少写floyed，但是这个算法解决这种所有顶点之间的关系很容易理解。在数据量不大的情况下是可以接受的。
该题思路：
①对于a胜于b，我们使得mp[a][b]=1.
②floyed更新各个点之间的关系。
③遍历每两对点，如果mp[i][j]==1或者mp[j][i]==1的话（i>j或者j>i的个数），计数加一。最后如果计数总数为n-1的话,就代表它和其他n-1个点的关系都确定，那么它的排名就确定。
代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;const int maxx=1e2+10;
int mp[maxx][maxx];
int n,m;inline void floyed()
{for(int k=1;k<=n;k++)//循环顺序！！！{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;}}}
}
inline void solve()
{int sum=0,ans=0;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(mp[i][j]||mp[j][i]) sum++;}if(sum==n-1) ans++;sum=0;}cout<<ans<<endl;
}
int main()
{memset(mp,0,sizeof(mp));scanf("%d%d",&n,&m);int x,y;while(m--){scanf("%d%d",&x,&y);mp[x][y]=1;}floyed();solve();return 0;
}

努力加油a啊，(^o)/~

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