poj 1258 -- Asteroids

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14426		Accepted: 7851

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).

 

思路：这道题就是求最小点覆盖，今天又重新预习了一遍离散数学，以前真对不起老师对不起政府。

 

而最小点覆盖=最大二分匹配。 所以可以用匈牙利算法求解。。

 

/*======================================================================
 *           Author :   kevin
 *         Filename :   Asteroids.cpp
 *       Creat time :   2014-07-13 08:51
 *      Description :
 ========================================================================*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define clr(a,b) memset(a,b,sizeof(a))
#define M 550

using namespace std;

int g[M][M];
int n,k,used[M],linker[M];
bool DFS(int u)
{
    for(int v = 0; v < n; v++){
        if(g[u][v] && !used[v]){
            used[v] = 1;
            if(linker[v] == -1 || DFS(linker[v])){
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}
int MaxMatch()
{
    int res = 0;
    clr(linker,-1);
    for(int u = 0; u < n; u++){
        clr(used,0);
        if(DFS(u)) res++;
    }
    return res;
}
int main(int argc,char *argv[])
{
    while(scanf("%d%d",&n,&k)!=EOF){
        int a,b;
        clr(g,0);
        for(int i = 0; i < k; i++){
            scanf("%d%d",&a,&b);
            g[a-1][b-1] = 1;
        }
        int ans = MaxMatch();
        printf("%d\n",ans);
    }
    return 0;
}