题干：

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).

题目大意：

在N*N的0和1组成的格子，一次操作可以将一行或一列的1全部变成0，问至少要进行多少次操作才能将所有的1全部变成0.

解题报告：

    套路建图方式、、、裸的最小点覆盖、、

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
vector<int> vv[MAX];
int nxt[MAX];
bool used[MAX];
int n,k;
bool find(int x) {
	int up = vv[x].size();
	for(int i = 0; i<up; i++) {
		int v = vv[x][i];
		if(used[v] == 1) continue;
		used[v] = 1;
		if(nxt[v] == -1 || find(nxt[v])) {
			nxt[v] = x;
			return 1;
		}
	}
	return 0;
}
int match() {
	int sum = 0;
	for(int i = 1; i<=n; i++) {
		memset(used,0,sizeof used);
		if(find(i)) sum++;
	}
	return sum;
}
int main()
{
	cin>>n>>k;
	memset(nxt,-1,sizeof nxt);
	for(int i = 1; i<=k; i++) {
		int x,y;
		scanf("%d%d",&x,&y);
		vv[x].pb(y);
	}
	printf("%d\n",match());

	return 0 ;
 }