tab lines clu roi rate map integer empty accept

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！

！

！！

主要是要懂得將問題轉化。

將行列轉化為二分兩個集合。然後坐標點（i，j）即化為i行j列可以匹配。最後找出最少的覆蓋點，由匈牙利定理：最少覆蓋點=最大匹配邊 。

假設還是不懂能夠參考這裏點擊打開

AC代碼例如以下：

#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;
}

POJ 3041 Asteroids