Asteroids||POJ3041

link:http://poj.org/problem?id=3041
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

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).

題解：
把行看成點集X，列看成點集Y，點（x，y）有行星就看做x和y之間連了一條邊，以此類推，最後所有點和邊構造了一個二分圖，如果點x0有一根邊，那麼與這個點相連的所有的邊都可以被一炮轟完，因爲這些邊對應的行星它們要麼同行要麼同列
這樣就是相當於求這個二分圖的最大匹配了，按模板一遍就過
AC代碼：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
int f[505];
bool fr[505][505];
bool vis[505];
int n;
int find(int x)
{
    int i;
    for ( i=1 ; i<=n ; i++ )
    {
        if ( fr[x][i] && !vis[i] )
        {
            vis[i] = true;
            if ( f[i]==-1 || find(f[i]) )
            {
                f[i] = x;
                return true;
            }

        }
    }
    return false;
}
int main()
{
    int k,i,ans;
    while(~scanf("%d%d",&n,&k))
    {
        ans=0;
        memset(f,-1,sizeof(f));
        memset(fr,false,sizeof(fr));
        for ( i=1 ; i<=k ; i++ )
        {
            int a,b;
            scanf("%d%d",&a,&b);
            fr[a][b] = true;
        }
        for ( i=1 ; i<=n ; i++)
        {
            memset(vis,false,sizeof(vis));
            if ( find(i) )
                ans++;
        }
        printf("%d\n",ans);
    }

return 0;
}