Asteroids （最小覆盖）

题目很简单，但是需要推到出二分图最大匹配 = 最小覆盖

最小覆盖：证明过程http://blog.sina.com.cn/s/blog_51cea4040100h152.html

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 41 11 32 23 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).

#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>using namespace std;int n;int map[502][502];bool visited[502];int match[502];bool find(int i)   ///查找当前的i是否可以匹配{    int j;    for(j=1;j<=n;j++)    {        if(map[i][j]&&!visited[j])        {            visited[j]=1;            if(match[j]==-1||find(match[j]))            {                match[j]=i;                return 1;            }        }    }    return 0;}int main(){    int k,i,x,y,ans;    while(~scanf("%d%d",&n,&k))    {        ans=0;        memset(map,0,sizeof(map));        memset(match,-1,sizeof(match));        for(i=0;i<k;i++)//对有意思的进行初始化        {            scanf("%d%d",&x,&y);            map[x][y]=1;        }        for(i=1;i<=n;i++)        {            memset(visited,0,sizeof(visited));//开始标记为全部没有访问            if(find(i))  ans++;        }        printf("%d\n",ans);    }    return 0;}