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 Python (3.4.3) (judge verdict: Accepted)

# -*- coding: utf-8 -*-
import heapq
import math

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = list(range(n))
    cycles = list(range(n, n-r, -1))
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

tmp = input()
N, M, R = [int(a) for a in tmp.split()]
tmp = input()
mokutekis = [int(a) for a in tmp.split()]
Rs = []
for i in range(M):
    tmp = input()
    Rs.append([int(a) for a in tmp.split()])

roots = {}
for a, b, dist in Rs:
    roots[(a, b)], roots[(b, a)] = dist, dist

minimums = {}
for i in range(1, N+1):
    minimums[(i, i)] = 0
for m in mokutekis:
    tmp_mins = {}
    merged = [False for i in range(N+1)]
    merged[m] = True
    merged_num = 1
    now = m
    while merged_num < N:
        minimum = [None, None]
        for i in range(1, N+1):
            if merged[i]:
                continue
            if (now, i) not in roots:
                new_dist = 100000*200+1
            else:
                new_dist = minimums[(m, now)] + roots[(now, i)]
            if (m, i) not in tmp_mins or tmp_mins[(m, i)] > new_dist:
                tmp_mins[(m, i)] = new_dist
                tmp_mins[(i, m)] = new_dist
            if minimum[0] is None or minimum[1] > tmp_mins[(m, i)]:
                minimum = i, tmp_mins[(m, i)]
        minimums[(m, minimum[0])], minimums[(minimum[0], m)] = minimum[1], minimum[1]
        now = minimum[0]
        merged[now] = True
        merged_num += 1

d_min = None
for perm in permutations(range(len(mokutekis))):
    perm = list(perm)
    d = 0
    d_max = None
    for p1, p2 in zip(perm, perm[1:]+[perm[0]]):
        m1, m2 = mokutekis[p1], mokutekis[p2]
        now_d = minimums[(m1, m2)]
        if d_max is None or d_max < now_d:
            d_max = now_d
        d += now_d
    d -= now_d
    if d_min is None or d_min > d:
        d_min = d
print(d_min)