poj 3041 Asteroids 最小顶点覆盖 && 二分匹配

Asteroids

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).

//把行作为第一点集，列作为第二点集，某行某列有障碍物时，那么连边，建图完毕。
//建图完成后，会发现此题明显是求最小顶点覆盖，又有最小顶点覆盖等于最大匹配，故用匈牙利算法求出最大匹配即可
//这是遇到的第二道最小顶点覆盖转最大匹配问题
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;

const int N = 510;

int match[N];
bool use[N];
bool s[N][N];
int n, k;

bool dfs(int v)
{
    for(int i = 1; i <= n; i++)
    {
        if(s[v][i] == true && use[i] == false)
        {
            use[i] = true;
            if(match[i] == -1 || dfs(match[i]))
            {
                match[i] = v;
                return true;
            }
        }
    }

    return false;
}

int hungary()
{
    int res = 0;
    memset(match, -1, sizeof match);
    for(int i = 1; i <= n; i++)
    {
        memset(use, 0, sizeof use);
        if(dfs(i)) res++;
    }

    return res;
}

int main()
{
    int a, b;

    while(~ scanf("%d%d", &n, &k))
    {
        memset(s, 0, sizeof s);
        for(int i = 0; i < k; i++)
        {
            scanf("%d%d", &a, &b);
            s[a][b] = true;
        }

        printf("%d\n", hungary());
    }

    return 0;
}