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)

#include <stdio.h>          // printf(), scanf()

#define MAX_V 15
#define min(a, b) ((a) < (b) ? (a) : (b))

static const int INF = 1000000000;

int d[MAX_V][MAX_V];
int dp[1 << MAX_V][MAX_V];
int V;

int
solve()
{
	int S, u, v;

	for (S = 0; S < (1 << V); ++S)
	{
		for (v = 0; v < V; ++v)
			dp[S][v] = INF;
	}

	dp[(1 << V) - 1][0] = 0;
	for (S = (1 << V) - 2; S >= 0; --S)
	{
		for (v = 0; v < V; ++v)
		{
			for (u = 0; u < V; ++u)
			{
				if (!(S >> u & 1))
				{
					int t = dp[S | (1 << u)][u] + d[v][u];
					dp[S][v] = min(dp[S][v], t);
				}
			}
		}
	}

	return dp[0][0];
}

int
main(int argc, char** argv)
{
	int E;
	int s, t, c;
	int i, j;

	scanf("%d %d", &V, &E);
	for (i = 0; i < V; ++i)
	{
		for (j = 0; j < V; ++j)
		{
			if (i == j)
				d[i][j] = 0;
			else
				d[i][j] = INF;
		}
	}

	for (i = 0; i < E; ++i)
	{
		scanf("%d %d %d", &s, &t, &c);
		d[s][t] = c;
	}


	int res = solve();
	if (res == INF)
		res = -1;

	printf("%d\n", res);
	return 0;
}