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

就是走路径的，用个二维数组，储存哪两个城市可以走，在用一个visit数组，储存走过的城市，然后dfs一下就好了；

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h>
#include<string.h>
#include<algorithm>
#define sf(a) scanf("%d",&a)
#define sff(a,b) scanf("%d%d",&a,&b)
#define sfff(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define pf printf
#define pb push_back
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod=1e9+100;
const db e=exp(1);
using namespace std;
const double pi=acos(-1.0);
int a[25][25],ans=1,visit[21];
int x,y,z;
vector<int>v;
void dfs(int n)
{if(v.size()==20)//存了20个说明走完了{if(a[v[19]][x]==1){pf("%d:  ",ans++);rep(i,0,v.size()){pf("%d ",v[i]);}pf("%d\n",x);}}rep(i,1,21){if(a[n][i]&&visit[i]){a[n][i]=0;visit[i]=0;//走过标记v.pb(i);//存入走过的城市，好在后面输出dfs(i);visit[i]=1;v.pop_back();a[n][i]=1;}}
}
int main()
{mm(a,0);rep(i,1,21){sfff(x,y,z);a[i][x]=1;a[i][y]=1;a[i][z]=1;visit[i]=1;}while(1){sf(x);if(!x) return 0;v.clear();//记得清空v.pb(x);visit[x]=0;dfs(x);}
}