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: Wrong Answer)

#include<stdio.h>
#define min(a,b) (a<b?a:b)
#define INF 10000000
int g[16][16],n,i,j,k,v,e,f,t,c;
int dp[1<<17][16];
int main()
{
	scanf("%d%d",&v,&e);
	for(i=0;i<v;i++)
		for(j=0;j<v;j++)
			g[i][j]=INF;
	for(i=1;i<(1<<16);i++)
		for(j=0;j<v;j++)
			dp[i][j]=INF;
	for(i=0;i<e;g[f][t]=c,i++)
		scanf("%d%d%d",&f,&t,&c);
	for(i=0;i<(1<<v);i++)
		for(j=0;j<v;j++)
			for(k=0;k<v;k++)
				if(g[j][k]<INF&&!(i&(1<<k)))
					dp[i+(1<<k)][k]=min(dp[i+(1<<k)][k],dp[i][j]+g[j][k]);
	printf("%d\n",dp[(1<<v)-1][0]==INF?-1:dp[(1<<v)-1][0]);
	return 0;
}