POJ 3660 Cow Contest

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6213 Accepted: 3360

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 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

 

题意：

给出牛的两两比赛结果 问有几头牛可以确定其排名

 

代码：

#include<iostream>#include<string>#include<cstring>#define MAX 102using namespace std;int map[MAX][MAX];int n,m;int main(){while(cin>>n>>m!=NULL){int a,b;memset(map,0,sizeof(map));for(int i=0;i<m;i++){cin>>a>>b;map[a][b]=1;}for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(map[i][k] && map[k][j])map[i][j]=1;int ans=0;for(int i=1;i<=n;i++){int tmp=0;for(int j=1;j<=n;j++)tmp+=(map[i][j]+map[j][i]);if(tmp==n-1)ans++;}cout<<ans<<endl;}return 0;}

思路：

map[i][j]表示i在j排名之前 则所有i能走到的点都在其排名之后 所有能走到i的点都在i排名之前

那么将所有i能走到的点 和所有能走到i的点相加若为(n-1)  则i的排名可以确定

中间用了类似于floy的方法来确定排名关系