POJ 3041--Asteroids【二分图 && 最小点数覆盖】

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17861		Accepted: 9729

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：

太空中有个N * N的网格，网格中分布了K个危险的小行星，有一种武器，只需要一枚子弹，就可以使某一行或者某一列上的所有小行星毁灭。问最少需要多少子弹，才能摧毁所有的小行星。

思路：

裸的最小点覆盖数题目。水水水。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 550
using namespace std;
int map[maxn][maxn];
int used[maxn];
int link[maxn];
int n, m;

void init(){
    memset(map, 0, sizeof(map));
}

void getmap(){
    while(m--){
        int a, b;
        scanf("%d%d", &a, &b);
        map[a][b] = 1;
    }
}

bool dfs(int x){
    for(int i = 1; i <= n; ++i){
        if(map[x][i] && !used[i]){
            used[i] = 1;
            if(link[i] == -1 || dfs(link[i])){
                link[i] = x;
                return true;
            }
        }
    }
    return false;
}

int hungary(){
    int ans = 0;
    memset(link, -1, sizeof(link));
    for(int i = 1; i <= n; ++i){
        memset(used, 0, sizeof(used));
        if(dfs(i))
            ans++;
    }
    return ans;
}

int main (){
    while(scanf("%d%d", &n, &m) != EOF){
        init();
        getmap();
        int sum = hungary();
        printf("%d\n", sum);
    }
    return 0;
}