N - Asteroids

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

  Hint

  INPUT DETAILS:
  The following diagram represents the data, where "X" is an asteroid and "." is empty space:
  X.X
  .X.
  .X.

OUTPUT DETAILS:
Bessie may fire across row 1 to destroy the asteroids at (1,1) and (1,3), and then she may fire down column 2 to destroy the asteroids at (2,2) and (3,2).

求发射多少次能将行星都摧毁，把坐标的横纵坐标可以当做是两个点集合，然后糗事求最小点覆盖，用匈牙利算法

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+100;
const double eps=1e-8;
using namespace std;
const double pi=acos(-1.0);
const int inf=0xfffffff;
const int N=505;
int pre[N];
int visit[N],line[N][N];
int n,m,y,x;
bool find(int x)
{
    rep(i,1,n+1)
    {
        if(line[x][i]&&visit[i]==0)
        {
            visit[i]=1;
            if(pre[i]==0||find(pre[i]))
            {
                pre[i]=x;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    while(~sf("%d%d",&n,&m))
    {
        mm(line,0);
        mm(pre,0);
        while(m--)
        {
            sf("%d%d",&x,&y);
            line[x][y]=1; 
        }
        int ans=0;
        rep(i,1,n+1)
        {
            mm(visit,0);
            if(find(i)) ans++;
        }
        
        pf("%d\n",ans);
    }
}