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++14 (judge verdict: Accepted)

#include <iostream>
using namespace std;
int n,m,a,b,c,d[15][15],dp[(1<<15)][15];
int main(void){
    cin>>n>>m;
    for(int i=0;i<15;i++){
        for(int j=0;j<15;j++){
            d[i][j]=-1;
        }
    }
    for(int i=0;i<m;i++){
        cin>>a>>b>>c;
        d[a][b]=c;
    }
    for(int i=0;i<(1<<n);i++){
        for(int j=0;j<15;j++){
            dp[i][j]=1e9;
        }
    }
    
    dp[0][0]=0;
    for(int i=0;i<(1<<n);i++){
        for(int k=0;k<n;k++){
            for(int j=0;j<n;j++){
                if(((i>>j)&1)==1)continue;
                if(d[k][j]!=-1){
                    dp[i|(1<<j)][j]=min(dp[i|(1<<j)][j],dp[i][k]+d[k][j]);
                }
            }
        }
    }
    if(dp[(1<<n)-1][0]==1e9)cout<<-1<<endl;
    else cout<<dp[(1<<n)-1][0]<<endl;
}