Poj Asteroids(最小顶点覆盖)

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17456		Accepted: 9498

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

如果这道题目，不放在训练记划，感觉那里也不像二分图。。。。。

看了讨论才知道，可以吧所有的行和列看成来集合，怪物所在的点，即为链接集合中的两个点的线段。故为最小点覆盖问题（最大匹配）

要尽量的发射子弹，即至少选择一个点来进行所有边的消除。

#include<iostream>
#include<cstring>
#include<cstdio>
bool linkmap[510][510],use[510];
int cropath[510],n,m;
int fi(int u)
{
    for(int i=1;i<=n;i++)
    {
        if(linkmap[u][i]&&!use[i])
        {
            use[i]=1;
            if(cropath[i]==-1||fi(cropath[i]))
            {
                cropath[i]=u;
                return -1;
            }
        }
    }
    return 0;
}
using namespace std;
int main()
{
    int a,b,c,k,i;
    ios::sync_with_stdio(false);
    while(cin>>n>>m)
    {
        memset(linkmap,false,sizeof(linkmap));
        memset(cropath,-1,sizeof(cropath));
        for(i=0;i<m;i++)
        {
            cin>>a>>b;
            linkmap[a][b]=true;
        }
        int ans=0;
        for(i=1;i<=n;i++)
        {
            if(fi(i))
            {
            memset(use,0,sizeof(use));
                ans++;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}