POJ 3041 Asteroids

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14371		Accepted: 7822

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).

匈牙利算法！

！

！！

主要是要懂得将问题转化。

将行列转化为二分两个集合。然后坐标点（i，j）即化为i行j列可以匹配。最后找出最少的覆盖点，由匈牙利定理：最少覆盖点=最大匹配边 。

假设还是不懂能够參考这里点击打开

AC代码例如以下：

#include<iostream>
#include<cstring>
using namespace std;

int v[505],s[505],map[505][505];
int n,m;

int xyl(int a)
{
    int i;
    for(i=1;i<=n;i++)
    {
        if(!v[i]&&map[a][i])
        {
            v[i]=1;
            if(!s[i]||xyl(s[i]))
            {
                s[i]=a;
                return 1;
            }
        }
    }
    return 0;
}

int main()
{
    int i,j;

    int a,b;
    while(cin>>n>>m)
    {
        memset(s,0,sizeof s);
        for(i=1;i<=m;i++)
        {
            cin>>a>>b;
            map[a][b]=1;
        }
        int sum=0;
        for(i=1;i<=n;i++)
        {
            memset(v,0,sizeof v);
            if(xyl(i))
                sum++;
        }
        cout<<sum<<endl;
    }
    return 0;
}