Problem:
Traveling Salesman Problem

For a given weighted directed graph G(V, E), find the distance of the shortest route that meets the following criteria:

It is a closed cycle where it ends at the same point it starts.

It visits each vertex exactly once.

Input

|V| |E|
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1

|V| is the number of vertices and |E| is the number of edges in the graph. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target vertices of i-th edge (directed) and di represents the distance between si and ti (the i-th edge).

Output

Print the shortest distance in a line. If there is no solution, print -1.

Constraints

2 ≤ |V| ≤ 15

0 ≤ di ≤ 1,000

There are no multiedge

Sample Input 1

4 6
0 1 2
1 2 3
1 3 9
2 0 1
2 3 6
3 2 4

Sample Output 1

16

Sample Input 2

3 3
0 1 1
1 2 1
0 2 1

Sample Output 2

-1

Solution in C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define repr(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define invrepr(i,a,b) for(int i=b-1;i>=a;i--)
#define invrep(i,n) invrepr(i,0,n)
#define repitr(itr,a) for(auto itr=a.begin();itr!=a.end();++itr)
const int MOD=1e9+7;


int main() {
    ios_base::sync_with_stdio(false); 

    int n,m;
    cin >> n >> m;
    vector<vector<int>> d(n,vector<int>(n,-1));
    rep(i,m) {
        int a,b,din,tin;
        cin >> a >> b >> din;
        d[a][b]=din;
    }
    vector<vector<int>> dp((1<<n),vector<int>(n,1e9));
    dp[0][0]=0;
    rep(bit,(1<<n)) {
        rep(i,n) {
            if (bit & (1<<i)) {
                rep(k,n) {
                    if (dp[bit-(1<<i)][k]!=1e9 && i!=k && d[k][i]>=0) {
                        dp[bit][i]=min(dp[bit][i],dp[bit-(1<<i)][k]+d[k][i]);
                    }
                }
            }
        }
    }
    int ans=1e9;
    ans=dp[(1<<n)-1][0];
    if (ans==1e9) ans=-1;
    cout << ans << endl;
    return 0;
}