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)

#define _CRT_SECURE_NO_WARNINGS
#define TRUE 1;
#define FALSE 0;

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>

int GetBit(int mask, int v){
	return (mask >> v) & 1;
}

int AddBit(int mask, int v){
	return mask | (1 << v);
}

int Task(int* pointRib, int n){
	int* dP = NULL;
	int maxMask = pow((double)2, (double)n);
	dP = (int*)calloc((maxMask)* n, sizeof(int));
	int mask = 0;
	int vertex = 0;
	short IO = TRUE;
	int i;
	for (mask = 1; mask < maxMask; mask++){
		for (vertex = 0; vertex < n; vertex++){
			for (i = 0; i < n; i++){
				if (1 != GetBit(mask, i) && 1 == GetBit(mask, vertex) && 0 != pointRib[vertex*n * 2 + i * 2] && (1 == mask || (1 < mask && 0 != dP[mask*n + vertex]))){
					dP[AddBit(mask, i)*n + i] = pointRib[vertex*n * 2 + i * 2 + 1] + dP[mask*n + vertex];
				}
			}
		}
	}
	int min = 0;
	for (vertex = 0; vertex < n; vertex++){
		if (0 != dP[(maxMask - 1)*n + vertex] && 0 != pointRib[vertex*n * 2]){
			if (0 == min || (min > (dP[(maxMask - 1)*n + vertex] + pointRib[vertex*n*2 + 1]))){
				min = dP[(maxMask - 1)*n + vertex] + pointRib[vertex*n*2 + 1];
			}
		}
	}
	return (0 != min) ? min : -1;
}

int main(){
	int n, m;
	scanf("%d %d", &n, &m);
	int i;
	int* rib = NULL;
	rib = calloc(n * n*2, sizeof(int));
	int x, y, weight;
	for (i = 0; i < m; i++){
		scanf("%d %d %d", &x, &y, &weight);
		rib[x*n * 2 + y * 2] = 1;
		rib[x*n * 2 + y * 2 + 1] = weight;
	}
	/*int j;
	int k;
	for (i = 0; i < n; i++){
		for (j = 0; j < n; j++){
			if (-1 == rib[i*n + j]){
				for (k = 0; k < n; k++){
					if (0 != rib[i*n + k]){
						rib[j*n + k] = rib[i*n + k];
					}
					if (0 != rib[j*n + k]){
						rib[i*n + k] = rib[j*n + k];
					}
				}
			}
		}
	}*/
	printf("%d\n", Task(rib, n));

	return 0;
}