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>
using namespace std;


int inf = 1000*16;
vector<vector<int>> dist(15, vector<int>(15, inf));  // グラフ構造
vector<vector<int>> dp(pow(2, 15), vector<int>(15, 0));

int rec(long long S, int v)
{
    if (S==0 && v==0) return 0;
    if (!(S & (1<<v))) return inf;

    if (dp[S][v]) return dp[S][v];
    int &ret = dp[S][v];
    ret = inf;
    for (int i=0; i<15; i++)
    {
        ret = min(ret, rec(S^(1<<v), i) + dist[i][v]);
    }
    return ret;
}

int main()
{
    int V, E;
    cin >> V >> E;

    // グラフの初期化
    for (int i=0; i<15; i++)
    {
        for (int j=0; j<15; j++) dist[i][j] = inf;
    }
    
    for (int i=0; i<E; i++)
    {
        int s,t,d;
        cin >> s >> t >> d;
        dist[s][t]=d;
    }

    int ans = rec((1<<V)-1, 0);
    if (ans >= inf) cout << -1 << endl;
    else cout << ans << endl;
}