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# (judge verdict: Accepted)

using System;
using System.Linq;

namespace DPL2_A
{
    class Program
    {
        const int INF = 1 << 30;
        const int MAX = 15;
        static int n;
        static int[,] d = new int[MAX, MAX];
        static int[,] dp = new int[1 << MAX, MAX];
        static void solve()
        {
            for (int i = 0; i < 1 << n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    dp[i, j] = INF;
                }
            }
            dp[(1 << n) - 1, 0] = 0;

            for (int S = (1 << n) - 2; S >= 0; S--)
            {
                for (int v = 0; v < n; v++)
                {
                    for (int u = 0; u < n; u++)
                    {
                        if ((S >> u & 1) == 0 && d[v,u] != INF)
                        {
                            dp[S, v] = Math.Min(dp[S, v], dp[S | 1 << u, u] + d[v, u]);
                        }
                    }
                }
            }
            Console.WriteLine(dp[0,0] == INF ? -1 : dp[0,0]);
        }
        static void Main(string[] args)
        {
            var t = scan;
            n = t[0];
            int e = t[1];

            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    d[i, j] = INF;
                }
            }
            for (int i = 0; i < e; i++)
            {
                var k = scan;
                d[k[0], k[1]] = k[2];
            }

            solve();
        }


        static int[] scan
        {
            get { return Console.ReadLine().Split().Select(int.Parse).ToArray(); }
        }
    }
}