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 Rust (1.15.1) (judge verdict: Time Limit Exceeded)


use std::vec::Vec;
use std::cmp::*;
use std::collections::BinaryHeap;
use std::cmp::Ordering;

pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T] where T: Ord {
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i-1]  {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i-1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }

    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }

        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] <= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();

        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j-1] < self[i-1]  {
            j -= 1;
        }

        // Step 4: Swap that element with the pivot
        self.swap(i-1, j);

        true
    }

}


#[derive(Copy, Clone, Eq, PartialEq)]
struct State {
    cost: usize,
    position: usize,
}

// The priority queue depends on `Ord`.
// Explicitly implement the trait so the queue becomes a min-heap
// instead of a max-heap.
impl Ord for State {
    fn cmp(&self, other: &State) -> Ordering {
        // Notice that the we flip the ordering on costs.
        // In case of a tie we compare positions - this step is necessary
        // to make implementations of `PartialEq` and `Ord` consistent.
        other.cost.cmp(&self.cost)
            //.then_with(|| self.position.cmp(&other.position))
    }
}

// `PartialOrd` needs to be implemented as well.
impl PartialOrd for State {
    fn partial_cmp(&self, other: &State) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

// Each node is represented as an `usize`, for a shorter implementation.
#[derive(Clone, Debug, Eq, PartialEq)]
struct Edge {
    node: usize,
    cost: usize,
}


#[derive(Clone,  Debug, Eq, PartialEq)]
struct Graph {
  adj_list: Vec<Vec<Edge>>,   // 隣接リスト
}

impl Graph {
  pub fn new(n: usize, v: &Vec<(usize, usize, usize)>) -> Graph {
    let mut list = vec![vec![]; n];
    for &(from, to, cost) in v {
      list[from].push(Edge {node: to, cost: cost});
      list[to].push(Edge {node: from, cost: cost});
    }

    Graph { adj_list: list, }
  }
  // Start at `start` and use `dist` to track the current shortest distance
  // to each node. This implementation isn't memory-efficient as it may leave duplicate
  // nodes in the queue. It also uses `usize::MAX` as a sentinel value,
  // for a simpler implementation.
  pub fn shortest_path(&self, start: usize, goal: usize) -> Option<usize> {
      // dist[node] = current shortest distance from `start` to `node`
      let mut dist: Vec<_> = (0..self.adj_list.len()).map(|_| std::usize::MAX).collect();

      let mut heap = BinaryHeap::new();

      // We're at `start`, with a zero cost
      dist[start] = 0;
      heap.push(State { cost: 0, position: start });

      // Examine the frontier with lower cost nodes first (min-heap)
      while let Some(State { cost, position }) = heap.pop() {
          // Alternatively we could have continued to find all shortest paths
          if position == goal { return Some(cost); }

          // Important as we may have already found a better way
          if cost > dist[position] { continue; }

          // For each node we can reach, see if we can find a way with
          // a lower cost going through this node
          for edge in &self.adj_list[position] {
              let next = State { cost: cost + edge.cost, position: edge.node };

              // If so, add it to the frontier and continue
              if next.cost < dist[next.position] {
                  heap.push(next);
                  // Relaxation, we have now found a better way
                  dist[next.position] = next.cost;
              }
          }
      }

      // Goal not reachable
      None
  }
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
  let (n, m, r) = {
    let v: Vec<usize> = read_vec();
    (v[0], v[1], v[2])
  };

  let mut v: Vec<usize> = read_vec::<usize>().iter().map(|&x| x - 1).collect();
  let mut t: Vec<(usize, usize, usize)> = Vec::new();
  for _ in 0..m {
    let s: Vec<usize> = read_vec();
    t.push((s[0] - 1, s[1] - 1, s[2]));
  }
  v.sort();
  let g = Graph::new(n, &t);
  let mut res = std::usize::MAX;
  loop {
    let mut tmp = 0;
    for i in 0..r-1 {
      match g.shortest_path(v[i], v[i+1]) {
        None => {
          tmp = std::usize::MAX;
          break;
        },
        Some(cost) => {
          tmp += cost;
        },
      }
    }

    res = min(res, tmp);
    if !v.next_permutation() {
      break;
    }
  }
  println!("{}", res);
}