poj3660 Cow Contest

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8967 Accepted: 5032

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

Source

USACO 2008 January Silver

floyd能过：

#include<stdio.h>#include<string.h>#include<algorithm>#define N 0x3f3f3f3fusing namespace std;int map[120][120],i,j,k,l,m,n,help,u,v;void floyd(){int i,j,k,l;for(i=1;i<=n;i++)for(j=1;j<=n;j++){//if(map[i][j]==N)//continue;for(k=1;k<=n;k++){if(map[j][i]+map[i][k]>1)//如果j能到达i并且i能到达k map[j][k]=1;//那么j就能到达k }}}int main(){while(scanf("%d%d",&n,&m)!=EOF){memset(map,0,sizeof(map));//先全部定义成0for(i=0;i<m;i++){scanf("%d%d",&u,&v); map[u][v]=1;//表示u能到达v }floyd();//for(i=1;i<=n;i++)//{//for(j=1;j<=n;j++)//printf("%d %d\n",map[i][j],map[j][i]);//printf("****\n");//}int ans=0;for(i=1;i<=n;i++){help=0;for(j=1;j<=n;j++)if(map[i][j]==1||map[j][i]==1)//计算当前这个点能到达的和能被到达的个数 help++;if(help==n-1)ans++;}printf("%d\n",ans);}}