# POJ 3041 Asteroids （最小点覆盖）

Asteroids

 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 21550 Accepted: 11708

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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
#define inf (1<<25)
using namespace std;

const int MAXN=510;
int uN,vN;//u,v数目
int g[MAXN][MAXN];
bool used[MAXN];
bool dfs(int u)//从左边开始找增广路径
{
int v;
for(v=0; v<vN; v++) //这个顶点编号从0开始，若要从1开始需要修改
if(g[u][v]&&!used[v])
{
used[v]=true;
{
//找增广路，反向
return true;
}
}
return false;//这个不要忘了，经常忘记这句
}
int hungary()
{
int res=0;
int u;
for(u=0; u<uN; u++)
{
memset(used,0,sizeof(used));
if(dfs(u))res++;
}
return res;
}
int main()
{
int k;
while(~scanf("%d%d",&uN,&k))
{
int r,c;
memset(g,0,sizeof(g));
vN=uN;
for(int i=0; i<k; i++)
{
scanf("%d%d",&r,&c);
g[r-1][c-1]=1;
}
printf("%d\n",hungary());
}
return 0;
}