poj 3660 Cow Contest

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5967 Accepted: 3219

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

 利用弗洛伊德传递关系。若要n个人比赛要确定某个人的排名则必须知道他与剩下的人的胜负关系

#include <stdio.h>#include<string.h>int maps[150][150],n;//map记录两只牛的比赛情况void floyd(){    int i,j,k;    for(k=1; k<=n; k++)        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)                if(maps[i][k]&&maps[k][j])//如果i牛打败了k牛且k牛打败了j牛                {                    maps[i][j]=1;//那么i牛也打败了j牛                    //printf("%d->%d %d\n",i,j,maps[i][j]);                }}int main(){    int m,i,j,a,b,sum,ans=0;    scanf("%d%d",&n,&m);    memset(maps,0,sizeof maps);    for(i=1; i<=m; i++)    {        scanf("%d%d",&a,&b);        maps[a][b]=1;    }    floyd();    for(i=1; i<=n; i++) //要判断一只牛的排名则必须它与剩下其它牛的胜负关系    {        sum=0;        for(j=1; j<=n; j++)            if(maps[i][j]||maps[j][i])//如果两只牛的关系可知                sum++;        if(sum==n-1)//如果和其余牛关系可知            ans++;    }    printf("%d\n",ans);    return 0;}