POJ

Cow Contest

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 12848
Accepted: 7149

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 cowA has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠ B), then cow A will always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM (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 andM
* 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

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题意：

牛在比赛，问知道两头牛的关系。最终可以确定多少个牛的名次

思路：

只要找到某只牛与其他所有牛的关系则其名次一定可以确定

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm> using namespace std;#define INF 0x3f3f3f3fint dis[105][105];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++)//i 可直接或间接打败 j if(dis[i][j] || (dis[i][k]&&dis[k][j]))dis[i][j]=1;}int main(){scanf("%d%d",&n,&m);memset(dis,0,sizeof(dis));for(int i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);dis[a][b]=1;//不能双向 }floyd();int sum=0;for(int i=1;i<=n;i++){int t=0;for(int j=1;j<=n;j++){if(dis[i][j] || dis[j][i])t++;}//只要确定了自己和其他所有的关系，则名次一定确定 if(t==n-1)sum++;}printf("%d\n",sum);return 0;}