Asteroids（二分图的最大分配）

最新推荐文章于 2023-02-26 12:07:16 发布

转载最新推荐文章于 2023-02-26 12:07:16 发布 · 141 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/guoyongzhi/p/3307704.html

本博客介绍了一个算法问题，目标是在一个N*N的网格中，使用一种可以清除整行或整列小行星的强大武器，找到消除所有K个小行星的最小射击次数。通过使用匈牙利算法实现的最大匹配策略，提供了详细的代码实现和样例输入输出说明。

~题目链接~

http://poj.org/problem?id=3041

~题目~

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12708		Accepted: 6918

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

Sample Output

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).

最小点覆盖=最大匹配算法匈牙利

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 const int maxn=510;
 5 using namespace std;
 6 
 7 int n;
 8 int map[maxn][maxn];
 9 int dis[maxn],vis[maxn];
10 
11 int BFS(int c)
12 {
13     int i;
14     for(i=1; i<=n; i++)
15     {
16         if(!vis[i] && map[c][i])
17         {
18             vis[i]=1;
19             if(dis[i]==-1 || BFS(dis[i]))
20             {
21                 dis[i]=c;
22                 return 1;
23             }
24         }
25     }
26     return 0;
27 }
28 int main()
29 {
30     int m;
31     while(~scanf("%d%d",&n,&m))
32     {
33         int a,b;
34         memset(map,0,sizeof(map));
35         memset(dis,-1,sizeof(dis));
36         while(m--)
37         {
38             cin>>a>>b;
39             map[a][b]=1;
40         }
41         int i,sum=0;
42         for(i=1; i<=n; i++)
43         {
44             memset(vis,0,sizeof(vis));
45             if(BFS(i))
46             sum++;
47         }
48         cout<<sum<<endl;
49     }
50 }

View Code

转载于:https://www.cnblogs.com/guoyongzhi/p/3307704.html