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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.*;
import java.math.*;
import java.util.*;

public class Main {
    private static boolean debug = false;
    private static boolean elapsed = false;

    private static PrintWriter _out = new PrintWriter(System.out);
    private static PrintWriter _err = new PrintWriter(System.err);

    private static class Road {
        int to;
        int cost;
        public Road(int to, int cost) {
            this.to = to;
            this.cost = cost;
        }
    }

    private List<List<Road>> roads = new ArrayList<>();
    private List<Integer> r = new ArrayList<>();

    private int ans = Integer.MAX_VALUE;

    private void solve(Scanner sc) {
        int N = sc.nextInt();
        for (int i = 0; i < N; ++i) {
            roads.add(new ArrayList<>());
        }
        int M = sc.nextInt();
        int R = sc.nextInt();
        for (int i = 0; i < R; ++i) {
            r.add(sc.nextInt() - 1);
        }
        for (int i = 0; i < M; ++i) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            int c = sc.nextInt();
            roads.get(a).add(new Road(b, c));
            roads.get(b).add(new Road(a, c));
        }

        search(new HashSet<>(), -1, 0);

        _out.println(ans);
    }
    private Set<String> memo = new HashSet<>();
    private void search(Set<Integer> set, int p, int d) {
        String key = set + "-" + p + "-" + d;
        if (memo.contains(key)) {
            return;
        }
        memo.add(key);

        if (set.size() >= r.size()) {
            if (d < ans) {
                ans = d;
            }
            return;
        }

        for (int t : r) {
            if (set.contains(t)) {
                continue;
            }
            if (r.contains(p)) {
                dr = Integer.MAX_VALUE;
                dr(new ArrayList<>(), p, t, -1, 0);
            } else {
                dr = 0;
            }
            set.add(t);
            search(set, t, d + dr);
            set.remove(t);
        }
    }
    private int dr;
    private Map<String, Integer> memo2 = new HashMap<>();
    private void dr(List<Integer> list, int f, int t, int p, int d) {
        String key = list + "-" + f + "-" + t + "-" + p + "-" + d;
        Integer tmp = memo2.get(key);
        if (tmp != null) {
            dr = tmp;
            return;
        }

        if (list.size() > 0 && list.get(list.size() - 1).equals(t)) {
            if (d < dr) {
                dr = d;
                memo2.put(key, dr);
            }
            return;
        }

        if (p < 0) {
            if (!list.contains(f)) {
                list.add(f);
                dr(list, f, t, f, d);
                list.remove(list.size() - 1);
            }
        } else {
            for (Road road : roads.get(p)) {
                if (!list.contains(road.to)) {
                    list.add(road.to);
                    dr(list, f, t, road.to, d + road.cost);
                    list.remove(list.size() - 1);
                }
            }
        }
    }
    private static BigInteger C(long n, long r) {
        BigInteger res = BigInteger.ONE;
        for (long i = n; i > n - r; --i) {
            res = res.multiply(BigInteger.valueOf(i));
        }
        for (long i = r; i > 1; --i) {
            res = res.divide(BigInteger.valueOf(i));
        }
        return res;
    }
    private static BigInteger P(long n, long r) {
        BigInteger res = BigInteger.ONE;
        for (long i = n; i > n - r; --i) {
            res = res.multiply(BigInteger.valueOf(i));
        }
        return res;
    }
    /*
     * 10^10 > Integer.MAX_VALUE = 2147483647 > 10^9
     * 10^19 > Long.MAX_VALUE = 9223372036854775807L > 10^18
     */
    public static void main(String[] args) {
        long S = System.currentTimeMillis();

        Scanner sc = new Scanner(System.in);
        new Main().solve(sc);
        _out.flush();

        long G = System.currentTimeMillis();
        if (elapsed) {
            _err.println((G - S) + "ms");
        }
        _err.flush();
    }
}