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: Wrong Answer)

import java.util.Map.Entry;
import java.io.*;
import java.util.*;

public class Main {
	
	public static void main(String[] args) throws IOException {
		FastScanner sc = new FastScanner(System.in);
		
		int N = sc.nextInt();
		int M = sc.nextInt();
		int R = sc.nextInt();
		int[] r = new int[R];
		for(int i = 0; i < R; i++) r[i] = sc.nextInt()-1;
		long[][] dist = new long[N][N];
		for(int i = 0; i < N; i++)
			for(int j = 0; j < N; j++) if(i != j) dist[i][j] = 20000001;
		for(int i = 0; i < M; i++){
			int A = sc.nextInt()-1;
			int B = sc.nextInt()-1;
			long C = sc.nextLong();
			dist[A][B] = C;
			dist[B][A] = C;
		}
		
		for(int k = 0; k < N; k++){
			for(int i = 0; i < N; i++){
				for(int j = 0; j < N; j++){
					dist[i][j] = Math.min(dist[i][j], dist[i][k]+dist[k][j]);
				}
			}
		}
		
		long min = Long.MAX_VALUE;
		do{
			long sum = 0;
			for(int i = 0; i < r.length-1; i++)
				sum += dist[r[i]][r[i+1]];
			min = Math.min(min, sum);
		}while(nextPermutation(r));
		
		System.out.println(min);
	}
	
	//配列版
	public static boolean nextPermutation(int[] a){
		for(int i = a.length - 1; i > 0; --i){
			if(a[i - 1] < a[i]){
				int swapIndex = find(a[i - 1], a, i, a.length - 1);
				int temp = a[swapIndex];
				a[swapIndex] = a[i - 1];
				a[i - 1] = temp;
				Arrays.sort(a, i, a.length);
				return true;
			}
		}
		return false;
	}

	//destより大きい要素の中で最小のもののインデックスを二分探索で探す
	private static int find(int dest, int[] a, int s, int e){
		if(s == e){
			return s;
		}
		int m = (s + e + 1) / 2;
		return a[m] <= dest ? find(dest, a, s, m - 1) : find(dest, a, m, e);
	}
}

class FastScanner {
    private InputStream in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    public FastScanner(InputStream in) {
		this.in = in;
	}
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() {
    	if (hasNextByte()) return buffer[ptr++];
    	else return -1;
    }
    private static boolean isPrintableChar(int c){
    	return 33 <= c && c <= 126;
    }
    public boolean hasNext() {
    	while(hasNextByte() && !isPrintableChar(buffer[ptr]))
    		ptr++; return hasNextByte();
    }
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDouble() {
    	return Double.parseDouble(next());
    }
}