poj 3660

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5695 Accepted: 3064

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：给出牛的个数N，M个结果A和B（A能打败B）。问有多少头牛的rank已经确定。

思路：求每一头牛相关的边数（即度的大小，如果度的大小等于n-1，表示其他牛与他的关系已经确定，则这个牛的rank也就确定了）。

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int n,m;int map[105][105];int degree[105];int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        int a,b;        memset(map,0,sizeof(map));        memset(degree,0,sizeof(degree));        for(int i=0; i<m; i++)        {            cin>>a>>b;            map[a-1][b-1]=1;        }        for(int k=0;k<n; k++)            for(int i=0; i<n; i++)                for(int j=0; j<n; j++)                    if(map[i][k]&&map[k][j])                        map[i][j]=1;        for(int i=0;i<n;i++)        for(int j=0;j<n;j++)        {            if(map[i][j])            {                degree[i]++;                degree[j]++;            }        }        int sum=0;        for(int i=0;i<n;i++)        if(degree[i]==n-1)        sum++;        cout<<sum<<endl;    }    return 0;}