POJ - 3041 Asteroids 【最小點覆寫】

Asteroids

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 25309	Accepted: 13659

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

思路：

點覆寫，在圖論中點覆寫的概念定義如下：對于圖G=(V,E)中的一個點覆寫是一個集合S⊆V使得每一條邊至少有一個端點在S中。

最小點覆寫=最大比對數

關于比對的一些概念：https://blog.csdn.net/flynn_curry/article/details/52966283

這裡的點表示x、y的坐标。最小點覆寫就是取最少的邊，能夠覆寫所有的點。

一個點表示為（x,y）的一條邊，直接跑匈牙利算法。

類似題：魔理沙又來偷書了

代碼：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define MAXN 510
int k,n,m;
int match[MAXN];
bool used[MAXN];
char G[MAXN][MAXN];
vector<int> vec[MAXN];
bool find(int u)
{
    for(int i=0;i<vec[u].size();i++)
    {
        int v=vec[u][i];
        if(!used[v])
        {
            used[v]=1;
            if(match[v]==-1 || find(match[v]))
            {
                match[v]=u;
                return true;
            }
        }
    }
    return false;
}
int hungry()
{
    int ans=0;
    memset(match,-1,sizeof match);
    for(int i=1;i<=n;i++)
    {
        memset(used,0,sizeof used);
        if(find(i)) ans++;
    }
    return ans;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<=n;i++)
            vec[i].clear();
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            vec[u].push_back(v);
        }
        printf("%d\n",hungry());
    }
    return 0;
}