Asteroids POJ - 3041 最小点覆盖集

Posted 2022-12-25 pengge666

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：

给一个N*N的矩阵，一些星球位于一些点上，要求消除所有星球，每次操作可消除一行或一列的星球，求最少要几次。

最小点覆盖定义：二分图中，选取最少的点数，使这些点和所有的边都有关联（把所有的边的覆盖），叫做最小点覆盖。

结论：最小点覆盖数 = 最大匹配数

思路：把横坐标作为集合X，纵坐标作为集合Y，对于点(x,y)，由X中的x连向Y中的y。对于每条边，只要x和y中有一个被删除即可，明显的最小点覆盖模型。

AC代码：

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<algorithm>
 5 using namespace std;
 6 #define maxn 5555
 7 int vis[maxn];
 8 int match[maxn];
 9 int e[maxn][maxn];
10 int n,m;
11 int dfs(int u)  //匈牙利算法模板
12     for(int i=1;i<=n;i++)
13         if(vis[i]==0&&e[u][i]==1)
14             vis[i]=1;
15             if(match[i]==0||dfs(match[i]))
16                 match[i]=u;
17                 return 1;
18             
19         
20     
21     return 0;
22 
23 int main()
24     cin>>n>>m;
25     for(int i=1;i<=m;i++)
26         int x,y;
27         cin>>x>>y;
28         e[x][y]=1;
29     
30     memset(match,0,sizeof(match));
31     int ans=0;
32     for(int i=1;i<=n;i++)
33         memset(vis,0,sizeof(vis));
34         if(dfs(i))
35             ans++;
36     
37     printf("%d\n",ans);
38     return 0;
39