1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <string>
 5 #include <cmath>
 6 #include <vector>
 7 #include <algorithm>
 8 using namespace std;
 9 #define INF 0x3f3f3f
10 int n, m;
11 int mp[11][11], dp[60000][11];
12 int fun(int pos){
13     return int(pow(3.0, pos));
14 }
15 int main(){
16     while(scanf("%d%d", &n, &m) != EOF){
17         int a, b, c;
18         for(int i = 1; i <= n; i++){
19             for(int j = 1; j <= n; j++){
20                 if(i == j) mp[i][j] = 0;
21                 else mp[i][j] = -1;
22             }
23         }
24         for(int i = 1; i <= m; i++){
25             scanf("%d%d%d", &a, &b, &c); //注意重边。 
26             if(mp[a][b] == -1) mp[a][b] = mp[b][a] = c;
27             else mp[a][b] = mp[b][a] = min(mp[a][b], c);
28         }
29         for(int i = 0; i < (fun(n)); i++){ 
30             for(int j = 1; j <= n; j++) dp[i][j] = -1;
31         } 
32         for(int pos = 1; pos <= n; pos++) dp[fun(pos-1)][pos] = 0;
33         
34         for(int i = 0; i < (fun(n)); i++){
35             for(int j = 1; j <= n; j++){
36                 if(dp[i][j] == -1) continue;
37                 for(int k = 1; k <= n; k++){
38                     if(mp[j][k] == -1) continue; //如果不连通 
39                     int tpos = i;
40                     for(int ii = 1; ii < k; ii++) tpos /= 3;
41                     tpos = tpos%3; //找到那一位。
42                     if(tpos == 2) continue;  //这个点已经走了两次了，不能再经过了 
43                     if(dp[i+fun(k-1)][k] == -1) dp[i+fun(k-1)][k] = dp[i][j] + mp[j][k];
44                     if(dp[i+fun(k-1)][k] >= dp[i][j] + mp[j][k]) dp[i+fun(k-1)][k] = dp[i][j] + mp[j][k];
45                 }
46             }
47         }
48         int sum = 0;
49         for(int i = 0; i < n; i++) sum += int(pow(3.0, i));
50         int Min = INF;
51         //sum是每个位上正好为1的状态，从sum开始循环。 
52         for(int i = sum; i < fun(n); i++){
53             for(int j = 1; j <= n; j++){
54                  bool flag = true;
55                 int temp = i, r;
56                 while(temp){
57                     r = temp%3;
58                     if(r == 0){
59                         flag = false; break;
60                     }
61                     temp /= 3;
62                 }
63                 if(flag == false) continue; //判断是否有位上为0，这样的状态是不满足条件的。 
64                 if(dp[i][j] <= Min && dp[i][j] != -1) Min = dp[i][j];
65             }
66         }
67         printf("%d\n",Min == INF?-1:Min);
68     }
69     return 0;
70 }