POJ 3041 Asteroids【二分图最大匹配.最小点覆盖】

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

 定理：
 1. 最小点权覆盖集=最小割=最大流
 2. 最大点权独立集=总权-最小点权覆盖集

相关证明：http://apps.hi.baidu.com/share/detail/16529351
分析：题意：给一个N*N的格点图，一枪可以干掉一排，问最少多少枪可以干掉所有敌人，可以把横坐标 i 纵坐标 j,看成是两个集合，求二分图最大匹配，即求最小点权覆盖。

code：

#include<stdio.h>
#include<string.h>
#define clr(x)memset(x,0,sizeof(x))
#define maxn 505
bool v[maxn];
bool g[maxn][maxn];
int link[maxn];
int n;
int find(int k)
{
    int i;
    for(i=1;i<=n;i++)
    {
        if(g[k][i]&&!v[i])
        {
            v[i]=1;
            if(link[i]==0||find(link[i]))
            {
                link[i]=k;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,k,p,q,tot;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        clr(g);  clr(link);
        while(k--)
        {
            scanf("%d%d",&p,&q);
            g[p][q]=1;
        }
        tot=0;
        for(i=1;i<=n;i++)
        {
            clr(v);
            if(find(i))
                tot++;
        }
        printf("%d\n",tot);
    }
    return 0;
}