【二分图最大匹配】poj3041

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.

* 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.

* Line 1: The integer representing the minimum number of times Bessie must shoot.

3 4
1 1
1 3
2 2
3 2

2

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

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).

大意：网格格子中有若干个星球，每次可以打掉一行或一列的所有星球，求打完所有点的最少次数。

方法：以行列建图，有星球的点行和列连一条边，求出最大匹配就是最少的次数

（最小的点覆盖 = 最大匹配数）

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const long N = 510;
bool a[N][N];
bool b[N];
long link[N];
long n, k;
bool find(long x)
{
    for(long i = 1; i <= n; i++)
    {
        if ((!b[i]) && (a[x][i]))
        {
            b[i] = true;
            if((link[i] == 0)||(find(link[i])))
            {
                link[i] = x;
                return true;
            }
        }
    }
    return false;
}
void init()
{
    memset(a, 0, sizeof(a));
    for (long i = 1; i <= k; i++)
    {
        long x, y;
        scanf("%d%d", &x, &y);
        a[x][y] = true;
    }
}
int main()
{
    freopen("poj3041.in", "r", stdin);

    while(scanf("%d%d", &n, &k)!=EOF)
    {
        init();
        long re = 0;
        memset(link, 0, sizeof(link));
        for(long i = 1; i <= n; i++)
        {
            memset(b, 0, sizeof(b));
            if (find(i)) re++;
        }
        cout << re <<endl;
    }

    return 0;
}