HDU2181(KB2-C)

哈密顿绕行世界问题

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3890    Accepted Submission(s): 2358

Problem Description

一个规则的实心十二面体，它的 20个顶点标出世界著名的20个城市，你从一个城市出发经过每个城市刚好一次后回到出发的城市。 

 

Input

前20行的第i行有3个数,表示与第i个城市相邻的3个城市.第20行以后每行有1个数m,m<=20,m>=1.m=0退出.

 

Output

输出从第m个城市出发经过每个城市1次又回到m的所有路线,如有多条路线,按字典序输出,每行1条路线.每行首先输出是第几条路线.然后个一个: 后列出经过的城市.参看Sample output

 

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: 5 1 2 3 4 8 7 17 18 14 15 16 9 10 11 12 13 20 19 6 5

2: 5 1 2 3 4 8 9 10 11 12 13 20 19 18 14 15 16 17 7 6 5

3: 5 1 2 3 10 9 16 17 18 14 15 11 12 13 20 19 6 7 8 4 5

4: 5 1 2 3 10 11 12 13 20 19 6 7 17 18 14 15 16 9 8 4 5

5: 5 1 2 12 11 10 3 4 8 9 16 15 14 13 20 19 18 17 7 6 5

6: 5 1 2 12 11 15 14 13 20 19 18 17 16 9 10 3 4 8 7 6 5

7: 5 1 2 12 11 15 16 9 10 3 4 8 7 17 18 14 13 20 19 6 5

8: 5 1 2 12 11 15 16 17 18 14 13 20 19 6 7 8 9 10 3 4 5

9: 5 1 2 12 13 20 19 6 7 8 9 16 17 18 14 15 11 10 3 4 5

10: 5 1 2 12 13 20 19 18 14 15 11 10 3 4 8 9 16 17 7 6 5

11: 5 1 20 13 12 2 3 4 8 7 17 16 9 10 11 15 14 18 19 6 5

12: 5 1 20 13 12 2 3 10 11 15 14 18 19 6 7 17 16 9 8 4 5

13: 5 1 20 13 14 15 11 12 2 3 10 9 16 17 18 19 6 7 8 4 5

14: 5 1 20 13 14 15 16 9 10 11 12 2 3 4 8 7 17 18 19 6 5

15: 5 1 20 13 14 15 16 17 18 19 6 7 8 9 10 11 12 2 3 4 5

16: 5 1 20 13 14 18 19 6 7 17 16 15 11 12 2 3 10 9 8 4 5

17: 5 1 20 19 6 7 8 9 10 11 15 16 17 18 14 13 12 2 3 4 5

18: 5 1 20 19 6 7 17 18 14 13 12 2 3 10 11 15 16 9 8 4 5

19: 5 1 20 19 18 14 13 12 2 3 4 8 9 10 11 15 16 17 7 6 5

20: 5 1 20 19 18 17 16 9 10 11 15 14 13 12 2 3 4 8 7 6 5

21: 5 4 3 2 1 20 13 12 11 10 9 8 7 17 16 15 14 18 19 6 5

22: 5 4 3 2 1 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5

23: 5 4 3 2 12 11 10 9 8 7 6 19 18 17 16 15 14 13 20 1 5

24: 5 4 3 2 12 13 14 18 17 16 15 11 10 9 8 7 6 19 20 1 5

25: 5 4 3 10 9 8 7 6 19 20 13 14 18 17 16 15 11 12 2 1 5

26: 5 4 3 10 9 8 7 17 16 15 11 12 2 1 20 13 14 18 19 6 5

27: 5 4 3 10 11 12 2 1 20 13 14 15 16 9 8 7 17 18 19 6 5

28: 5 4 3 10 11 15 14 13 12 2 1 20 19 18 17 16 9 8 7 6 5

29: 5 4 3 10 11 15 14 18 17 16 9 8 7 6 19 20 13 12 2 1 5

30: 5 4 3 10 11 15 16 9 8 7 17 18 14 13 12 2 1 20 19 6 5

31: 5 4 8 7 6 19 18 17 16 9 10 3 2 12 11 15 14 13 20 1 5

32: 5 4 8 7 6 19 20 13 12 11 15 14 18 17 16 9 10 3 2 1 5

33: 5 4 8 7 17 16 9 10 3 2 1 20 13 12 11 15 14 18 19 6 5

34: 5 4 8 7 17 18 14 13 12 11 15 16 9 10 3 2 1 20 19 6 5

35: 5 4 8 9 10 3 2 1 20 19 18 14 13 12 11 15 16 17 7 6 5

36: 5 4 8 9 10 3 2 12 11 15 16 17 7 6 19 18 14 13 20 1 5

37: 5 4 8 9 16 15 11 10 3 2 12 13 14 18 17 7 6 19 20 1 5

38: 5 4 8 9 16 15 14 13 12 11 10 3 2 1 20 19 18 17 7 6 5

39: 5 4 8 9 16 15 14 18 17 7 6 19 20 13 12 11 10 3 2 1 5

40: 5 4 8 9 16 17 7 6 19 18 14 15 11 10 3 2 12 13 20 1 5

41: 5 6 7 8 4 3 2 12 13 14 15 11 10 9 16 17 18 19 20 1 5

42: 5 6 7 8 4 3 10 9 16 17 18 19 20 13 14 15 11 12 2 1 5

43: 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5

44: 5 6 7 8 9 16 17 18 19 20 1 2 12 13 14 15 11 10 3 4 5

45: 5 6 7 17 16 9 8 4 3 10 11 15 14 18 19 20 13 12 2 1 5

46: 5 6 7 17 16 15 11 10 9 8 4 3 2 12 13 14 18 19 20 1 5

47: 5 6 7 17 16 15 11 12 13 14 18 19 20 1 2 3 10 9 8 4 5

48: 5 6 7 17 16 15 14 18 19 20 13 12 11 10 9 8 4 3 2 1 5

49: 5 6 7 17 18 19 20 1 2 3 10 11 12 13 14 15 16 9 8 4 5

50: 5 6 7 17 18 19 20 13 14 15 16 9 8 4 3 10 11 12 2 1 5

51: 5 6 19 18 14 13 20 1 2 12 11 15 16 17 7 8 9 10 3 4 5

52: 5 6 19 18 14 15 11 10 9 16 17 7 8 4 3 2 12 13 20 1 5

53: 5 6 19 18 14 15 11 12 13 20 1 2 3 10 9 16 17 7 8 4 5

54: 5 6 19 18 14 15 16 17 7 8 9 10 11 12 13 20 1 2 3 4 5

55: 5 6 19 18 17 7 8 4 3 2 12 11 10 9 16 15 14 13 20 1 5

56: 5 6 19 18 17 7 8 9 16 15 14 13 20 1 2 12 11 10 3 4 5

57: 5 6 19 20 1 2 3 10 9 16 15 11 12 13 14 18 17 7 8 4 5

58: 5 6 19 20 1 2 12 13 14 18 17 7 8 9 16 15 11 10 3 4 5

59: 5 6 19 20 13 12 11 10 9 16 15 14 18 17 7 8 4 3 2 1 5

60: 5 6 19 20 13 14 18 17 7 8 4 3 10 9 16 15 11 12 2 1 5

 

Author

Zhousc

 

Source

ECJTU 2008 Summer Contest

 

裸的dfs。

 //2017-03-01
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;

 int M[][], ans[], vis[], m, kase;

 void dfs(int city, int step)
 {
     if(step ==  && city == m){
         cout<<++kase<<":  ";
         for(int i = ; i < ; i++)
               cout<<ans[i]<<" ";
         cout<<m<<endl;
         return;
     }
     ans[step] = city;
     for(int i = ; i < ; i++)
     {
         if(!vis[M[city][i]])
         {
             vis[M[city][i]] = ;
             dfs(M[city][i], step+);
             vis[M[city][i]] = ;
         }
     }
 }

 int main()
 {
     while(cin>>M[][]>>M[][]>>M[][])
     {
         sort(M[], M[]+);
         for(int i = ; i <= ; i++)
         {
             cin>>M[i][]>>M[i][]>>M[i][];
             sort(M[i], M[i]+);
         }
         while(cin>>m && m)
         {
             memset(vis, , sizeof(vis));
             kase = ;
             dfs(m, );
         }
     }

     return ;
 }