Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.

Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.

Now, they will play a game of tag as follows:

1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.

2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.

3. Go back to step 1.

Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.

Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.

It can be proved that the game is bound to end.

Constraints

2 \leq N \leq 10^5

1 \leq u,v \leq N

u \neq v

1 \leq A_i,B_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N u v
A_1 B_1
:
A_{N-1} B_{N-1}

Output

Print the number of moves Aoki will perform before the end of the game.

Sample Input 1

5 4 1
1 2
2 3
3 4
3 5

Sample Output 1

2

If both players play optimally, the game will progress as follows:

Takahashi moves to Vertex 3.

Aoki moves to Vertex 2.

Takahashi moves to Vertex 5.

Aoki moves to Vertex 3.

Takahashi moves to Vertex 3.

Here, Aoki performs two moves.

Note that, in each move, it is prohibited to stay at the current vertex.

Sample Input 2

5 4 5
1 2
1 3
1 4
1 5

Sample Output 2

1

Sample Input 3

2 1 2
1 2

Sample Output 3

0

Sample Input 4

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

Sample Output 4

5

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

//! ----------------------------------------------
//! Framework <https://github.com/vain0x/procon>
//!
//! See the bottom of file for solution.
//! ----------------------------------------------

#![allow(unused_imports)]
#![allow(non_snake_case)]

use std::cell::RefCell;
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Display, Formatter, Write as FmtWrite};
use std::io::{stderr, stdin, BufRead, Write};
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;

/// Print values to standard error if debug mode.
#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), stderr());
            writeln!(err, "\x1B[33m{}\x1B[0m = {:?}", e, $e).unwrap()
        })*
    };
}

/// Read from standard input and parse each word.
/// - `read!(T, U, ..)` parses a line as a tuple of words.
/// - `read![[T]]` parses a line as an array of words.
/// - `read![..; N]` parses `N` lines, using `read!(..)` repeatedly.
#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

/// Read a line from standard input.
#[allow(dead_code)]
fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();

    #[allow(deprecated)]
    buf.trim_right().to_owned()
}

// -----------------------------------------------
// Solution
// -----------------------------------------------

// 鬼(青木)の初期位置を根とする根つき木を計算
// 兎(高橋)は最も深いリーフに逃げ込みたい
// リーフを深い順に調べる
// 兎の初期位置とリーフのLCAを求める
// 兎と鬼がLCAを目指すとき鬼が先に到達するならダメ
// 兎が先に到達するなら兎がそのリーフに潜って追いつめられるまでをシミュレーションして解とする

struct Solver {
    N: usize,
    // 兎の初期位置
    U: usize,
    // 鬼の初期位置
    V: usize,
    G: Vec<Vec<usize>>,
    // 深さ、端点
    leaves: Vec<(usize, usize)>,
    // a[u][h]: u の 2^h 世代上の親要素
    ancestors: Vec<Vec<usize>>,
    depths: Vec<usize>,
}

impl Solver {
    fn dfs(&mut self, u: usize, parent: usize, depth: usize) {
        self.depths[u] = depth;
        debug!((u, parent, depth));

        let mut h = 0;
        // u の 2^h 世代上
        let mut a = parent;
        while a < self.N {
            self.ancestors[u].push(a);
            debug!((u, h, a));

            if h >= self.ancestors[a].len() {
                break;
            }

            // u の 2^(h+1) 世代上は a の 2^h 世代上
            a = self.ancestors[a][h];
            h += 1;
        }

        let mut is_leaf = true;

        for i in 0..self.G[u].len() {
            let v = self.G[u][i];
            if v == parent {
                continue;
            }

            is_leaf = false;
            self.dfs(v, u, depth + 1);
        }

        if is_leaf {
            debug!(("leaf", u));
            self.leaves.push((depth, u));
        }
    }

    fn lca(&self, mut u: usize, mut v: usize) -> usize {
        debug!(("lca", u, v));

        while u != v {
            let ud = self.depths[u];
            let vd = self.depths[v];

            if ud == vd {
                let mut h = 0;
                while h + 1 < self.ancestors[u].len()
                    && self.ancestors[u][h + 1] != self.ancestors[v][h + 1]
                {
                    h += 1;
                }

                debug!(("ud==vd", u, h, self.ancestors[u][h]));
                debug!(("ud==vd", v, h, self.ancestors[v][h]));
                u = self.ancestors[u][h];
                v = self.ancestors[v][h];
                continue;
            }

            if ud > vd {
                std::mem::swap(&mut u, &mut v);
            }
            assert!(ud < vd);

            let mut h = 0;
            while h + 1 < self.ancestors[v].len() && (1 << (h + 1)) < vd - ud {
                h += 1;
            }

            debug!(("ud<vd", v, h, self.ancestors[v][h]));
            v = self.ancestors[v][h];
        }

        debug!(("  -->", u));
        u
    }

    fn solve(&mut self) -> usize {
        let (V, N) = (self.V, self.N);
        self.dfs(V, N, 0);

        self.leaves.sort();
        self.leaves.reverse();

        let ud = self.depths[self.U];

        for &(yd, y) in self.leaves.iter() {
            let x = self.lca(self.U, y);
            let xd = self.depths[x];

            assert!(ud >= xd);
            assert!(yd >= xd);

            // ud - xd : 兎の x に至る移動回数
            // xd : 鬼の x に至る移動回数
            if ud - xd >= xd {
                // 鬼が先に x に到達するので兎は y にいけない
                debug!(("ogre", yd, y, xd, x, ud - xd, xd));
                continue;
            }

            // 兎が y に至る移動回数
            let s = (ud - xd) + (yd - xd);

            // 鬼の移動回数
            let mut t = s - min(s, 1);

            // 兎が初めから y にいたのでなければ鬼が先手
            if s != 0 {
                t += 1;
            }

            // 鬼が兎を追いつめるまでの鬼の移動回数を加算
            t += (xd - (ud - xd)) / 2;

            debug!(("rabbit", yd, y, xd, x, s, t));
            return t;
        }

        debug!("???");
        0
    }
}

fn main() {
    let (N, U, V) = read!(usize, usize, usize);
    let T = read![usize, usize; N - 1];

    let mut G = vec![vec![]; N];
    for (u, v) in T {
        G[u - 1].push(v - 1);
        G[v - 1].push(u - 1)
    }

    let k = Solver {
        N: N,
        U: U - 1,
        V: V - 1,
        G: G,
        leaves: vec![],
        ancestors: vec![vec![]; N],
        depths: vec![0; N],
    }
    .solve();

    println!("{}", k)
}