| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5989 | Accepted: 3234 |
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 5
4 3
4 2
3 2
1 2
2 5
Sample Output
2
Source
//============================================================================
// Name : POJ.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
/*
* floyed算法，传递闭包。如果一个点和其余点的关系都是确定的，则这个的排名是确定的
*/
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
const int MAXN=;
int win[MAXN][MAXN];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)==)
{
memset(win,,sizeof(win));
int u,v;
while(m--)
{
scanf("%d%d",&u,&v);
win[u][v]=;
}
for(int k=;k<=n;k++)
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
if(win[i][k]&&win[k][j])
win[i][j]=;
int ans=;
int j;
for(int i=;i<=n;i++)
{
for(j=;j<=n;j++)
{
if(i==j)continue;
if(win[i][j]==&&win[j][i]==)break;//关系不确定
}
if(j>n)ans++;
}
printf("%d\n",ans);
}
return ;
}