poj---3660

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

题意：有n头牛比赛，m种比赛结果，最后问你一共有多少头牛的排名被确定了，其中如果a战胜b，b战胜c，则也可以说a战胜c，即可以传递胜负。求能确定排名的牛的数目。
分析：
如果一头牛被x头牛打败，打败y头牛，且x+y=n-1，则我们容易知道这头牛的排名就被确定了，所以我们只要将任何两头牛的胜负关系确定了，在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<stdio.h>#include<math.h>#include<algorithm>#include<string.h>#include<stdlib.h>#define N 110using namespace std;int maps[N][N];int n, m;void floyd(){     for(int k = 1; k <= n; k++)     {         for(int i = 1; i <= n; i++)         {             for(int j = 1; j <= n; j++)             {                 if(maps[i][k] && maps[k][j])                    maps[i][j]=1;             }         }     } } int main() {     while(scanf("%d%d",&n,&m)!=EOF)     {         memset(maps, 0, sizeof(maps));         int u, v;         for(int i=1;i<=m;i++)         {             scanf("%d %d",&u,&v);             maps[u][v]=1;         }         floyd();         int sol=0;         for(int i=1; i<=n; i++)         {             int cnt=n-1;             for(int j=1;j<=n;j++)             {                 if(maps[j][i] || maps[i][j])                    cnt--;             }             if(cnt==0)                sol++;         }         printf("%d\n", sol);     }     return 0; }