POJ

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13569 Accepted: 7556

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：每头牛都有一个能力值，能力高的会打败能力低的牛，给出n头牛和m次两两比赛的结果，问如果按能力值从小到大排序，可以确定出几头牛的位次。

思路：能够确定出位次的牛，一定与其他n-1只牛都有固定的关系。因为牛数很少，所以用floyd松弛，然后遍历一遍，找出所有与其他牛都存在关系的牛的个数。

#include <iostream>#include <cstdio>#include <map>#include <cstring>#define ll long longusing namespace std;int n,m;int mp[200][200];int main(){while(cin>>n>>m){memset(mp,0,sizeof(mp));while(m--){int x,y;cin>>x>>y;mp[x][y]=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]==1&&mp[k][j]==1){mp[i][j]=1;}}}}int ans=0;for(int i=1;i<=n;i++){int cnt=0;for(int j=1;j<=n;j++){if(i==j)continue;if(mp[i][j]==1||mp[j][i]==1)cnt++;}if(cnt>=n-1){ans++;}}printf("%d\n",ans);}}