poj 3041 Asteroids (最小点覆盖)

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 · 76 阅读

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原文链接：http://www.cnblogs.com/GetcharZp/p/9466419.html

在一片危险的陨石带中，Bessie必须使用她的飞船武器清除所有障碍。通过消除每一行或每一列的陨石，她可以穿越这片区域。任务是找到清除所有陨石所需的最少射击次数。

Asteroids

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26239		Accepted: 14182

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

C/C++：

 1 #include <map>
 2 #include <queue>
 3 #include <cmath>
 4 #include <vector>
 5 #include <string>
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <climits>
 9 #include <iostream>
10 #include <algorithm>
11 #define INF 0xffffff
12 using namespace std;
13 const int my_max = 505;
14 
15 int n, k, a, b, my_ans, my_line[my_max][my_max], my_book[my_max], my_right[my_max];
16 
17 bool my_dfs(int x)
18 {
19     for (int i = 1; i <= n; ++ i)
20     {
21         if (!my_book[i] && my_line[x][i])
22         {
23             my_book[i] = 1;
24             if (!my_right[i] || my_dfs(my_right[i]))
25             {
26                 my_right[i] = x;
27                 return true;
28             }
29         }
30     }
31     return false;
32 }
33 
34 int main()
35 {
36     while (~scanf("%d%d", &n, &k))
37     {
38         my_ans = 0;
39         memset(my_line, 0, sizeof(my_line));
40         memset(my_right, 0, sizeof(my_right));
41 
42         while (k --)
43         {
44             scanf("%d%d", &a, &b);
45             my_line[a][b] = 1;
46         }
47         for (int i = 1; i <= n; ++ i)
48         {
49             memset(my_book, 0, sizeof(my_book));
50             if (my_dfs(i)) my_ans ++;
51         }
52 
53         printf("%d\n", my_ans);
54     }
55     return 0;
56 }

转载于:https://www.cnblogs.com/GetcharZp/p/9466419.html