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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

class PriorityQueue
  MIN = -(1 << 60)

  attr_reader :size

  def initialize(capacity = nil, &block)
    @size = 0
    @heap = if capacity.nil?
      Array.new
    else
      # capacity には、queue のサイズとしてありえる最大の数を指定する想定。
      # @heap[1] から使用するため、配列サイズを 1つ余分に確保する。
      Array.new(capacity + 1)
    end
    @get_key = block || :itself.to_proc
  end

  def empty?
    size == 0
  end

  def pop
    return nil if @size <= 0
    max = @heap[1]
    @heap[1] = @heap[@size]
    @size -= 1
    max_heapify(1)
    max
  end

  def push(x)
    @size += 1
    i = @size
    @heap[i] = x
    while i > 1 && (@get_key.call(@heap[i / 2]) <=> @get_key.call(@heap[i])) < 0
      t = @heap[i / 2]
      @heap[i / 2] = @heap[i]
      @heap[i] = t
      i = i / 2
    end
  end

  def top
    return nil if @size <= 0
    @heap[1]
  end

  def to_a
    return [] if @size == 0
    @heap[1..@size]
  end

  private

  def max_heapify(i)
    l = 2 * i
    r = 2 * i + 1
    largest =
      l <= @size && (@get_key.call(@heap[l]) <=> @get_key.call(@heap[i])) > 0 ? l : i
    largest = r \
      if r <= @size && (@get_key.call(@heap[r]) <=> @get_key.call(@heap[largest])) > 0
    unless largest == i
      t = @heap[i]
      @heap[i] = @heap[largest]
      @heap[largest] = t
      max_heapify(largest)
    end
  end
end

N, M, R = gets.split.map(&:to_i)
r_list = gets.split.map { |a| a.to_i - 1}
graph = Array.new(N) { [] }
M.times do
  u, v, c = gets.split.map(&:to_i)
  u -= 1
  v -= 1
  graph[u].push([v, c])
  graph[v].push([u, c])
end

def shortest_path(graph, s)
  w_list = Array.new(graph.size, 1 << 60)
  w_list[s] = 0
  queue = PriorityQueue.new do |_, c|
    -c
  end
  queue.push([s, 0])
  while !queue.empty?
    u, w = queue.pop
    next if w > w_list[u]
    graph[u].each do |v, c|
      if w_list[u] + c < w_list[v]
        new_cost = w_list[u] + c
        w_list[v] = new_cost
        queue.push([v, new_cost])
      end
    end
  end
  w_list
end

d = N.times.map do |i|
  shortest_path(graph, i)
end

ans = 1 << 60
r_list.permutation.each do |l|
  ret = 0
  u = l[0]
  l.each do |v|
    ret += d[u][v]
    u = v
  end
  ans = ret if ret < ans
end
puts ans