poj 3041 Asteroids （最大匹配最小顶点覆盖——匈牙利模板题）

http://poj.org/problem?id=3041

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12601		Accepted: 6849

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

 

【题解】：

　　最大匹配最小顶点覆盖

 

【code】：

/**
Judge Status:Accepted     Memory:1880K    
Time:16MS        Language:G++    
Code Lenght:1182B   Author:cj
*/

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>

#define N 550
using namespace std;

int n;
int map[N][N];
int cx[N],cy[N],mark[N];

int path(int u)
{
    int j;
    for(j=1;j<=n;j++)
    {
        if(map[u][j]&&!mark[j])
        {
            mark[j]=1;
            if(cy[j]==-1||path(cy[j]))
            {
                cx[u] = j;
                cy[j] = u;
                return 1;
            }
        }
    }
    return 0;
}

int maxMatch()
{
    memset(cx,-1,sizeof(cx));
    memset(cy,-1,sizeof(cy));
    int i;
    int res = 0;
    for(i=1;i<=n;i++)
    {
        if(cx[i]==-1)
        {
            memset(mark,0,sizeof(mark));
            res+=path(i);
        }
    }
    return res;
}

int main()
{
    int k,i;
    scanf("%d%d",&n,&k);
    memset(map,0,sizeof(map));
    for(i=1;i<=k;i++)
    {
        int r,c;
        scanf("%d%d",&r,&c);
        map[r][c]=1;
    }
    printf("%d\n",maxMatch());
    return 0;
}