HDU 2181 哈密顿绕行世界问题 （dfs）

Sample Input

2 5 20

1 3 12

2 4 10

3 5 8

1 4 6

5 7 19

6 8 17

4 7 9

8 10 16

3 9 11

10 12 15

2 11 13

12 14 20

13 15 18

11 14 16

9 15 17

7 16 18

14 17 19

6 18 20

1 13 19

5

0

Sample Output

 1 #include <iostream>
 2
 3 using namespace std;
 4
 5 int Map[21][21] = { 0 };  //邻接矩阵
 6 int path[21];              //记录每一次答案的路线
 7 bool visited[21] = { false };
 8 int num = 1; //路径的编号
 9
10
11 void dfs(int city, int cur) //city为当前访问的城市，已经访问了cur个城市
12 {
13     path[cur] = city;    //将当前城市记录在路径上
14     visited[city] = true;
15
16     if (cur == 20)        //如果已经访问了20个城市
17     {
18         if (Map[city][path[1]] == 1)    //这个城市与起点城市之间存在路径
19         {
20             cout << num << ":  ";
21             num++;
22             for (int i = 1; i <= 20; ++i)
23                 cout << path[i] << ‘ ‘;
24             cout << path[1] << endl;
25         }
26     }
27
28     for (int i = 1; i <= 20; ++i)
29     {
30         if (Map[i][city] == 1 && !visited[i])
31         {
32             dfs(i, cur+1);        //注意此处不能是++cur
33             visited[i] = false;    //dfs完要把访问过的顶点重新置为未访问状态
34         }
35     }
36
37 }
38
39
40 int main()
41 {
42     //创建邻接矩阵
43     int city;
44     for (int i = 1; i <= 20; ++i)
45     {
46         for (int j = 1; j <= 3; ++j)
47         {
48             cin >> city;
49             Map[i][city] = 1;
50         }
51     }
52
53     int m;
54     while (cin >> m && m != 0)
55     {
56         dfs(m, 1);
57         num = 1;
58     }
59
60
61     return 0;
62
63 }

原文地址：https://www.cnblogs.com/FengZeng666/p/10352790.html