Problem:
Score : 400 points

Problem Statement

There are N towns in the State of Atcoder, connected by M bidirectional roads.

The i-th road connects Town A_i and B_i and has a length of C_i.

Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order).

She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road.

If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?

Constraints

2≤N≤200

1≤M≤N×(N-1)/2

2≤R≤min(8,N) (min(8,N) is the smaller of 8 and N.)

r_i≠r_j (i≠j)

1≤A_i,B_i≤N, A_i≠B_i

(A_i,B_i)≠(A_j,B_j),(A_i,B_i)≠(B_j,A_j) (i≠j)

1≤C_i≤100000

Every town can be reached from every town by road.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M R
r_1 ... r_R
A_1 B_1 C_1
:
A_M B_M C_M

Output

Print the distance traveled by road if Joisino visits the towns in the order that minimizes it.

Sample Input 1

3 3 3
1 2 3
1 2 1
2 3 1
3 1 4

Sample Output 1

2

For example, if she visits the towns in the order of 1, 2, 3, the distance traveled will be 2, which is the minimum possible.

Sample Input 2

3 3 2
1 3
2 3 2
1 3 6
1 2 2

Sample Output 2

4

The shortest distance between Towns 1 and 3 is 4. Thus, whether she visits Town 1 or 3 first, the distance traveled will be 4.

Sample Input 3

4 6 3
2 3 4
1 2 4
2 3 3
4 3 1
1 4 1
4 2 2
3 1 6

Sample Output 3

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <math.h>
#include <queue>
#include <set>
#include <stack>
#include <set>
#include <tuple>

using namespace std;
typedef vector<int>VI;
typedef long long LL;

#define IOSWORK ios::sync_with_stdio(0); cin.tie(0)
#define FOR(x, b, e) for(int x = b; x<= (e); x++)
#define FORD(x, b, e) for(int x = b; x>= (e); x--)
#define REP(x, n) for(int x = 0; x<(n); ++x)
#define ALL(c) (c).begin(), (c).end()
#define SIZE(x) ((int)(x).size())
#define VAR(v, n) typeof(n) v = (n)
#define FOREACH(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define SCAN_INT(n) scanf("%i",&n)
#define SCAN_P_INT(n,m) scanf("%i %i",&n,&m)
#define PRINT_P_INT(n,m) printf("%i %i\n",n,m);
#define PRINT_INT(n) printf("%i\n",n)

#define PB push_back
#define ST first
#define ND second
#define INF 1000000000


#define MAXSIZE 210
int n,m,r;
int suma = INF;
int odleglosci[MAXSIZE][MAXSIZE];
VI wierzcholki;
int zajete[10];

void permutacje(VI &ciag, int i){

    if(i==r-1){
        int temp_sum=0;
        REP(j,SIZE(ciag)-1)
          temp_sum+=odleglosci[ciag[j]][ciag[j+1]];//printf("%i ",odleglosci[wierzcholki[ciag[j]]] [wierzcholki[ciag[j+1]]]);

        suma=min(temp_sum,suma);
    }
    else{
        FOR(j,i,r-1){
            swap(ciag[i],ciag[j]);
            permutacje(ciag,i+1);
            swap(ciag[i],ciag[j]);
        }
    }

}

int main(){
    SCAN_P_INT(n,m);
    SCAN_INT(r);

    REP(i,r){
        int temp;
        SCAN_INT(temp);
        wierzcholki.PB(temp);
    }

    FOR(i,1,n)
        FOR(j,1,n){
            if(i==j)
                odleglosci[i][j]=0;
            else
                odleglosci[i][j]=INF;
        }

    REP(i,m){
        int a,b,c;
        scanf("%i %i %i",&a,&b,&c);
        odleglosci[a][b]=c;
        odleglosci[b][a]=c;
    }

    FOR(i,1,n)
        FOR(j,1,n)
            FOR(k,1,n){
                odleglosci[j][k]=min(odleglosci[j][k],odleglosci[j][i]+odleglosci[i][k]);
        }

    permutacje(wierzcholki,0);


    PRINT_INT(suma);
    return 0;
}