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)

import sys
stdin = sys.stdin

def li(): return map(int, stdin.readline().split())
def li_(): return map(lambda x: int(x)-1, stdin.readline().split())
def lf(): return map(float, stdin.readline().split())
def ls(): return stdin.readline().split()
def ns(): return stdin.readline().rstrip()
def lc(): return list(ns())
def ni(): return int(stdin.readline())
def nf(): return float(stdin.readline())

from itertools import permutations
from heapq import heappush, heappop

def dijkstra(graph:list, node:int, start:int) -> list:
    # 未探索のノードは距離INF
    INF = float("inf")
    dist = [INF]*node
    
    # 始点ノードの距離を0とし、dfsのためのpriority queを作成
    dist[start] = 0
    heap = [(0,start)]
    searched = [False for _ in range(node)]
    
    # 未探索のノードをpriority queueに入れる
    while heap:
        cost, cur_node = heappop(heap)
        
        if searched[cur_node] or cost > dist[cur_node]:
            continue
        
        for nex_cost, nex_node in graph[cur_node]:
            dist_cand = dist[cur_node] + nex_cost
            if dist_cand < dist[nex_node]:
                dist[nex_node] = dist_cand
                heappush(heap, (dist[nex_node], nex_node))
    
        searched[cur_node] = True
        
        if sum(searched) == node:
            break
    
    return dist

n,m,r = li()
r = list(li_())
graph = [[] for _ in range(n)]
for _ in range(m):
    a,b,c = li()
    a -= 1
    b -= 1
    graph[a].append((c,b))
    graph[b].append((c,a))
    
dists = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n):
    dists[i] = dijkstra(graph, n, i)

routes = list(permutations(r))

ans = 2**64-1
for route in routes:
    temp = 0
    for i in range(len(route)-1):
        temp += dists[route[i]][route[i+1]]
        
    ans = min(ans,temp)
    
print(ans)