POJ 3041 Asteroids 【匈牙利算法最小点覆盖】

原题链接：http://poj.org/problem?id=3041

题目来源：http://acm.hust.edu.cn:8080/judge/contest/view.action?cid=16967#problem/E

算法：匈牙利算法（二分匹配，求最小点覆盖数）

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10768		Accepted: 5823

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).

Source

USACO 2005 November Gold

题意：给出一个N*N的矩阵，有的位置有行星。已知一枪可以击溃一行或一列上的所有行星。问最少多少枪可以击溃       所有的行星。

思路：以行作为X集合，以列作为Y集合,一个行星在(x,y)，则x对应X中的点向y对应Y中的点连一条边，则某个顶点一       旦被选，则与之相连的边（也就是行星）都会被选，也就是选出最少的顶点覆盖所有的边，即最小顶点覆盖。

算法：

      Konig定理：一个二分图中的最大匹配数等于这个图中的最小点覆盖数。

      最小点覆盖数：假如选中了一个点就相当于覆盖了以它为端点的所有边，你需要选择最少的点覆盖所有的边。

PS：模板题目，直接套用匈牙利算法求最大的匹配数的模板即可。关键是理解如何建图。

#include<stdio.h>
#include<string.h>
const int maxn=510;
int g[maxn][maxn];
int linker[maxn];
bool used[maxn];
int n,m;
int uN,vN;
int find(int u)
{
    int v;
    for(v=1;v<=vN;v++)
    {
        if(!used[v] && g[u][v])
        {
            used[v]=true;
            if(linker[v]==-1 || find(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    }
    return false;
}
int hugary()
{
    int u;
    int sum=0;
    memset(linker,-1,sizeof(linker));
    for(u=1;u<=uN;u++)
    {
        memset(used,false,sizeof(used));
        if(find(u))   sum++;
    }
    return sum;
}
int main()
{
    int u,v;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        uN=vN=n;
        memset(g,0,sizeof(g));
        while(m--)
        {
            scanf("%d%d",&u,&v);
            g[u][v]=1;
        }
        printf("%d\n",hugary());
    }
}