Problem:
Score : 400 points

Problem Statement

You are given a tree with N vertices.

Here, a tree is a kind of graph, and more specifically, a connected undirected graph with N-1 edges, where N is the number of its vertices.

The i-th edge (1≤i≤N-1) connects Vertices a_i and b_i, and has a length of c_i.

You are also given Q queries and an integer K. In the j-th query (1≤j≤Q):

find the length of the shortest path from Vertex x_j and Vertex y_j via Vertex K.

Constraints

3≤N≤10^5

1≤a_i,b_i≤N (1≤i≤N-1)

1≤c_i≤10^9 (1≤i≤N-1)

The given graph is a tree.

1≤Q≤10^5

1≤K≤N

1≤x_j,y_j≤N (1≤j≤Q)

x_j≠y_j (1≤j≤Q)

x_j≠K,y_j≠K (1≤j≤Q)

Input

Input is given from Standard Input in the following format:

N
a_1 b_1 c_1
:
a_{N-1} b_{N-1} c_{N-1}
Q K
x_1 y_1
:
x_{Q} y_{Q}

Output

Print the responses to the queries in Q lines.

In the j-th line j(1≤j≤Q), print the response to the j-th query.

Sample Input 1

5
1 2 1
1 3 1
2 4 1
3 5 1
3 1
2 4
2 3
4 5

Sample Output 1

3
2
4

The shortest paths for the three queries are as follows:

Query 1: Vertex 2 → Vertex 1 → Vertex 2 → Vertex 4 : Length 1+1+1=3

Query 2: Vertex 2 → Vertex 1 → Vertex 3 : Length 1+1=2

Query 3: Vertex 4 → Vertex 2 → Vertex 1 → Vertex 3 → Vertex 5 : Length 1+1+1+1=4

Sample Input 2

7
1 2 1
1 3 3
1 4 5
1 5 7
1 6 9
1 7 11
3 2
1 3
4 5
6 7

Sample Output 2

5
14
22

The path for each query must pass Vertex K=2.

Sample Input 3

10
1 2 1000000000
2 3 1000000000
3 4 1000000000
4 5 1000000000
5 6 1000000000
6 7 1000000000
7 8 1000000000
8 9 1000000000
9 10 1000000000
1 1
9 10

Sample Output 3

17000000000

Solution in Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import scala.collection.mutable
import scala.collection.mutable.ArrayBuffer
import scala.io.StdIn._
object Main extends App {

  val n = readInt()
  val nodes = Array.fill(n){Nil: List[Edge]}
  for (_ ← 1 until n) {
    val Array(a, b, c) = readLine.trim.split(' ').map(_.toInt)
    nodes(a - 1) ::= Edge(b - 1, c)
    nodes(b - 1) ::= Edge(a - 1, c)
  }
  val parents = Array.fill(n){ArrayBuffer[Int]()}
  val queue = mutable.Queue[Int](0)
  val depth = Array.fill(n){Int.MaxValue}
  depth(0) = 0
  val length = Array.fill(n){0L}
  while (queue.nonEmpty) {
    val current = queue.dequeue()
    for (Edge(to, cost) ← nodes(current) if depth(to) > depth(current)) {
      parents(to).append(current)
      while(parents(to).length <= parents(parents(to).last).length) {
        parents(to).append(parents(parents(to).last)(parents(to).length - 1))
      }
      depth(to) = depth(current) + 1
      length(to) = length(current) + cost
      queue.enqueue(to)
    }
  }
  val Array(q, k) = readLine.trim.split(' ').map(_.toInt)
  for (_ ← 0 until q) {
    val Array(x, y) = readLine.trim.split(' ').map(_.toInt - 1)
    val ca = mlca(x, k - 1)
    val cb = mlca(y, k - 1)
    println(length(x) + length(k - 1) * 2 + length(y) - length(ca) * 2 - length(cb) * 2)
  }

  def mlca(a: Int, b: Int): Int = {
    if (a == b) a
    else if (depth(a) == depth(b)){
      val first = findFirst(0, parents(a).length){i ⇒
        parents(a)(i) == parents(b)(i)
      }
      if (first == 0) parents(a).head
      else mlca(parents(a)(first - 1), parents(b)(first - 1))
    }else if (depth(a) > depth(b)) {
      val first = findFirst(0, parents(a).length){i ⇒
        depth(parents(a)(i)) <= depth(b)
      }
      if (first == 0) mlca(parents(a).head, b)
      else mlca(parents(a)(first - 1), b)
    }else {
      val first = findFirst(0, parents(b).length){i ⇒
        depth(parents(b)(i)) <= depth(a)
      }
      if (first == 0) mlca(a, parents(b).head)
      else mlca(a, parents(b)(first - 1))
    }
  }

  def findFirst(from: Int, until: Int)(@inline predicate: Int ⇒ Boolean): Int = {
    var left = from
    var right = until
    while(left < right){
      val mid = (left + right) / 2
      if (predicate(mid)) right = mid
      else left = mid + 1
    }
    right
  }

  case class Edge(to: Int, cost: Long)
}