poj 3041 二分图最小点集覆盖

          二分图最小点集覆盖==二分图最大匹配，，知道了这个公式，这道题就很水了。。。。。。。题目：

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8487		Accepted: 4532

Description

Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice points of the grid. 

Fortunately, Bessie has a powerful weapon that can vaporize all the asteroids in any given row or column of the grid with a single shot.This weapon is quite expensive, so she wishes to use it sparingly.Given the location of all the asteroids in the field, find the minimum number of shots Bessie needs to fire to eliminate all of the asteroids.

Input

* Line 1: Two integers N and K, separated by a single space. 
* Lines 2..K+1: Each line contains two space-separated integers R and C (1 <= R, C <= N) denoting the row and column coordinates of an asteroid, respectively.

Output

* Line 1: The integer representing the minimum number of times Bessie must shoot.

Sample Input

3 4
1 1
1 3
2 2
3 2

Sample Output

2

ac代码：

#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#include <cstdio>
using namespace std;
vector<int> ss[505];
int flag[505],visted[505];
bool dfs(int x){
	for(int i=0;i<ss[x].size();++i){
		if(!visted[ss[x][i]]){
		  visted[ss[x][i]]=1;
		  if(!flag[ss[x][i]]||dfs(flag[ss[x][i]])){
		    flag[ss[x][i]]=x;
			return true;
		  }
		}
	}
	return false;
}
int main(){
	
	int n,m;
  while(~scanf("%d%d",&n,&m)){
    memset(ss,0,sizeof(ss));
	memset(flag,0,sizeof(flag));
	
	int x,y;
	while(m--){
	  scanf("%d%d",&x,&y);
	  ss[x].push_back(y);
	}
	int sum=0;
	for(int i=1;i<=n;++i){
		memset(visted,0,sizeof(visted));
	  if(dfs(i))
		  sum++;
	}
	printf("%d\n",sum);
  }
  return 0;
}