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

use std::io::{BufRead, Write};
use std::io;
use std::cmp::Ordering;
use std::collections::BinaryHeap;
use std::usize;

#[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.
        match other.cost.cmp(&self.cost) {
            ::std::cmp::Ordering::Equal => self.position.cmp(&other.position),
            x => x,
        }
    }
}

// `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)]
struct Edge {
    node: usize,
    cost: usize,
}

// Dijkstra's shortest path algorithm.

// 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.
fn shortest_path(adj_list: &Vec<Vec<Edge>>, start: usize, goal: usize) -> Option<usize> {
    // dist[node] = current shortest distance from `start` to `node`
    let mut dist: Vec<_> = (0..adj_list.len()).map(|_| 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 &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 solve<R: BufRead, W: Write>(mut input: R, mut output: W) {
    let mut s = String::new();
    input.read_line(&mut s).unwrap();
    let n: usize = s.trim().parse().unwrap();
    let mut adjacency = vec![Vec::<Edge>::new(); n];
    for _ in 0..n - 1 {
        s.clear();
        input.read_line(&mut s).unwrap();
        let (a, b, c): (usize, usize, usize) = {
            let mut split = s.split_whitespace();
            let a = split.next().unwrap().parse().unwrap();
            let b = split.next().unwrap().parse().unwrap();
            let c = split.next().unwrap().parse().unwrap();
            (a, b, c)
        };
        adjacency[b - 1].push(Edge {
            node: a - 1,
            cost: c,
        });
        adjacency[a - 1].push(Edge {
            node: b - 1,
            cost: c,
        });
    }
    s.clear();
    input.read_line(&mut s).unwrap();
    let (q, k): (usize, usize) = {
        let mut split = s.split_whitespace();
        let a = split.next().unwrap().parse().unwrap();
        let b = split.next().unwrap().parse().unwrap();
        (a, b)
    };
    for _ in 0..q {
        s.clear();
        input.read_line(&mut s).unwrap();
        let (x, y): (usize, usize) = {
            let mut split = s.split_whitespace();
            let a = split.next().unwrap().parse().unwrap();
            let b = split.next().unwrap().parse().unwrap();
            (a, b)
        };
        writeln!(
            output,
            "{}",
            shortest_path(&adjacency, x - 1, k - 1).unwrap() +
                shortest_path(&adjacency, k - 1, y - 1).unwrap()
        ).unwrap();
    }
}

fn main() {
    let stdin = io::stdin();
    let stdout = io::stdout();
    solve(stdin.lock(), stdout.lock());
}

#[cfg(test)]
mod tests {
    fn test(input: &str, output: &str) {
        let mut v: Vec<u8> = Vec::new();
        ::solve(input.as_bytes(), &mut v);
        assert_eq!(String::from_utf8(v).unwrap(), output);
    }

    #[test]
    fn case1() {
        test(
            "5
1 2 1
1 3 1
2 4 1
3 5 1
3 1
2 4
2 3
4 5\n",
            "3
2
4\n",
        );
    }
    #[test]
    fn case2() {
        test(
            "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\n",
            "5
14
22\n",
        );
    }
    #[test]
    fn case3() {
        test(
            "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\n",
            "17000000000\n",
        );
    }
}