hdu 1530 Maximum Clique (最大包)

最新推荐文章于 2021-05-01 00:01:29 发布

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原文链接：http://www.cnblogs.com/GetcharZp/p/9462826.html

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本文探讨了在给定图中寻找最大团的问题，详细解释了一种递归深度优先搜索算法，通过遍历所有可能的顶点组合来找出包含最多顶点的完全子图。文章提供了完整的C/C++代码实现，展示了如何检查子图是否构成团，并通过递归调用逐步构建最大团。

Maximum Clique
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5380 Accepted Submission(s): 2776

Problem Description
Given a graph G(V, E), a clique is a sub-graph g(v, e), so that for all vertex pairs v1, v2 in v, there exists an edge (v1, v2) in e. Maximum clique is the clique that has maximum number of vertex.

Input
Input contains multiple tests. For each test:

The first line has one integer n, the number of vertex. (1 < n <= 50)

The following n lines has n 0 or 1 each, indicating whether an edge exists between i (line number) and j (column number).

A test with n = 0 signals the end of input. This test should not be processed.

Output
One number for each test, the number of vertex in maximum clique.

Sample Input
5
0 1 1 0 1
1 0 1 1 1
1 1 0 1 1
0 1 1 0 1
1 1 1 1 0
0

Sample Output
4

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 = 55;
14 
15 int n, my_map[my_max][my_max], my_ans, my_set[my_max];
16 
17 bool my_is_clique(int my_depth, int my_point)
18 {
19     for (int i = 1; i < my_depth; ++ i)
20         if (!my_map[my_set[i]][my_point])
21             return false;
22     return true;
23 }
24 
25 void my_dfs(int my_depth, int my_point)
26 {
27     //if (my_depth + n - my_point + 1 <= my_ans) return;
28     for (int i = my_point; i <= n; ++ i)
29     {
30         if (my_is_clique(my_depth + 1, i))
31         {
32             my_set[my_depth + 1] = i;
33             my_dfs(my_depth + 1, i + 1);
34         }
35     }
36     if (my_depth > my_ans) my_ans = my_depth;
37 }
38 
39 int main()
40 {
41     while (~scanf("%d", &n), n)
42     {
43         memset(my_map, 0, sizeof(my_map));
44         memset(my_set, 0, sizeof(my_set));
45 
46         for (int i = 1; i <= n; ++ i)
47             for (int j = 1; j <= n; ++ j)
48                 scanf("%d", &my_map[i][j]);
49         my_ans = 0;
50         my_dfs(0, 1);
51         printf("%d\n", my_ans);
52     }
53     return 0;
54 }