POJ-3041 Asteroids,二分匹配解决棋盘问题。

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K

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).

题意：N*N的棋盘上有一些障碍，有一种武器能一次清除一行或者一列上的障碍，给出这些障碍的坐标，求最少需要用多少次才能将障碍全部清除。

思路：又到了我们的模型建立阶段，题意已经很清晰了；

    主要是构图：

    将每一行当成一个点，构成集合1，

    每一列也当成一个点，构成集合2；

    每一个障碍物的位置坐标将集合1与集合2中的点连接起来，也就是将每一个障碍物作为连接节点的边。

    这样可以轻易的得出本题是一个最小点覆盖的问题，假设1个行节点覆盖了5个列节点，

    即这个行节点与这5个列节点间有5条边（即五个障碍物），

    由于这5条边都被那个行节点覆盖，即表明这5个障碍物都在同一列上，

    于是可以一颗炸弹全部清除，而本题也就转化成求最小点覆盖数的问题，即求最大二分匹配。

    

    模型建立好以后，此题就变成了一个裸模板题了。

//const int INF=0x3f3f3f3f;
const int N=500+10;
int n,m,w[N][N],linked[N],v[N];
int dfs(int u)
{
    for(int i=1;i<=n;i++)
    if(!v[i]&&w[u][i])
        {
            v[i]=1;
            if(linked[i]==-1||dfs(linked[i]))
            {
                linked[i]=u;
             return 1;
            }
        }
    return 0;
}
void hungary()
{
    int ans=0;
    memset(linked,-1,sizeof(linked));
    for(int i=1;i<=n;i++)
    {
        memset(v,0,sizeof(v));
        if(dfs(i)) ans++;
    }
    printf("%d\n",ans);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int u,v;
        memset(w,0,sizeof(w));
        while(m--)
        {
            scanf("%d%d",&u,&v);
            w[u][v]=1;
        }
        hungary();
    }
    return 0;
}

此题让我拍手一声：好题。我想，算法的魅力就在于此吧。