poj3041—Asteroids（最小点覆盖）

传送门：点我

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).

大意：

输入N，k，表示在N*N的地图中有k个陨石，以下k行是陨石的横纵坐标，开一枪能让一排或一列的陨石消失，求最少开几枪能让所有陨石消失。

思路：x，y坐标相连可看成一条边，让所有的坐标都消失，转化为让所有边消失，即最小顶点覆盖数=最大匹配数

上马：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
#define N 510
vector<int>q[N];
int match[N];
bool vis[N];
bool hungary(int u)
{
    int i;
    for(i=0; i<q[u].size(); i++)
    {
        int v=q[u][i];
        if(!vis[v])
        {
            vis[v]=1;
            if(!match[v]||hungary(match[v]))
            {
                match[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int n,k,i;
    while(~scanf("%d%d",&n,&k))
    {
        for(i=1; i<=n; i++)
            q[i].clear();
        int a,b;
        for(i=1; i<=k; i++)
        {
            scanf("%d%d",&a,&b);
            q[a].push_back(b);
        }
        memset(match,0,sizeof(match));
        int ans=0;
        for(i=1; i<=n; i++)
        {
            memset(vis,0,sizeof(vis));
            ans+=hungary(i);
        }
        printf("%d\n",ans);
    }
    return 0;
}