[floyd]poj3660 Cow Contest

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

给出各个牛之间的强弱关系，判断多少个牛的排名能被确定下来。
用prim做，对于i,k和j，如果i>k且k>j则能说明i>j，修改所有的传递关系。
在数组内做完预处理后，对每个牛判断与各个的排名关系。如果这头牛和其他牛之间都有对应的胜负关系，那么就说明它的排名可以确定，加入计数器，统计完所有的牛以后再输出。

#include <queue>#include <iostream>#include<cstdio>#include<cstring>using namespace std;const int MAXN=102;int d[MAXN][MAXN];int N,M;int main(){    int a,b;    while(cin>>N>>M)    {        memset(d,0,sizeof(d));        for(int i = 0; i < M; i++)        {            cin>>a>>b;            d[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(d[i][k] && d[k][j])                        d[i][j]=1;        int ans = 0;        for(int i = 1; i<=N;i++)        {            int res = N-1;            for(int j = 1;j<=N;j++)            if(d[i][j] || d[j][i])                res--;            if (!res)                ans++;        }        cout<<ans<<endl;    }}