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 Python3 (judge verdict: Accepted)

def main():
  V, E = map(int, input().split())
  INF = float('INF')
  graph = [[] for _ in range(V)]
  for i in range(E):
    s, t, d = map(int, input().split())
    graph[s].append([t, d])
  dp = [[INF] * V for _ in range(1<<V)]
  dp[(1<<V)-1][0] = 0
  for bit in range((1<<V)-2, -1, -1):
    for v in range(V):
      for u, d in graph[v]:
        if not (bit>>u & 1):
          dp[bit][v] = min(dp[bit|1<<u][u]+d, dp[bit][v])

  if(dp[0][0] == INF):
    print(-1)
  else:
    print(dp[0][0])
  
if __name__ == '__main__':
	main()