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比较简单的dfs题目代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>const int maxn=1e5+5;
typedef long long ll;
using namespace std;int Map[25][25];
int m;
int vis[25];
int dis[25];
int cnt;
void dfs(int x,int step)
{dis[step]=x;if(step==20&&Map[x][m]==1){printf("%d:  ",cnt++);for(int t=1;t<=20;t++){printf("%d ",dis[t]);}printf("%d\n",m);return;}for(int t=1;t<=20;t++){if(Map[x][t]==1&&vis[t]==0){vis[t]=1;dfs(t,step+1);vis[t]=0;}}}
int main()
{int x;for(int t=1;t<=20;t++){for(int j=0;j<3;j++){scanf("%d",&x);Map[t][x]=1;}}while(cin>>m){if(m==0){break;}else{cnt=1;vis[m]=1;dfs(m,1);vis[m]=0;}}return 0;
}