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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.Set;

public class Main implements Runnable {
    public static void main(String[] args) {
        new Thread(null,new Main(), "" ,Runtime.getRuntime().maxMemory()).start();
    }

    // メイン処理
    public void run() {
        FastScanner sc = new FastScanner();
        int n = sc.nextInt();
        int u = sc.nextInt() - 1;
        int v = sc.nextInt() - 1;
        Map<Integer, Set<Integer>> routeMap = new HashMap<>(n);
        for(int i = 0;i < n - 1;i++) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            Set<Integer> fromA = routeMap.get(Integer.valueOf(a));
            if(fromA == null) fromA = new HashSet<>();
            fromA.add(Integer.valueOf(b));
            routeMap.put(Integer.valueOf(a), fromA);
            Set<Integer> fromB = routeMap.get(Integer.valueOf(b));
            if(fromB == null) fromB = new HashSet<>();
            fromB.add(Integer.valueOf(a));
            routeMap.put(Integer.valueOf(b), fromB);
        }

        Deque<int[]> que = new ArrayDeque<>();
        Map<Integer, Set<Integer>> possibility1 = new HashMap<>(n);
        int[] start = {u, 0};
        que.offer(start);
        Set<Integer> startSet = new HashSet<>();
        startSet.add(Integer.valueOf(u));
        possibility1.put(Integer.valueOf(0), startSet);
        for(int i = 1;i <= 2*n;i++) {
            int[] here = que.poll();
            Set<Integer> route = routeMap.get(Integer.valueOf(here[0]));
            Set<Integer> nextPossibility = possibility1.get(Integer.valueOf(here[1] + 1));
            if(nextPossibility == null) nextPossibility = new HashSet<>();
            for(Integer next : route) {
                nextPossibility.add(next);
                int[] tmp = {next.intValue(), here[1] + 1};
                que.add(tmp);
            }
            possibility1.put(Integer.valueOf(here[1] + 1), nextPossibility);
        }

        que.clear();
        Map<Integer, Set<Integer>> possibility2 = new HashMap<>(n);
        int[] start2 = {v, 0};
        que.offer(start2);
        Set<Integer> startSet2 = new HashSet<>();
        startSet2.add(Integer.valueOf(v));
        possibility2.put(Integer.valueOf(0), startSet2);
        for(int i = 1;i <= 2*n;i++) {
            int[] here = que.poll();
            Set<Integer> route = routeMap.get(Integer.valueOf(here[0]));
            Set<Integer> nextPossibility = possibility2.get(Integer.valueOf(here[1] + 1));
            if(nextPossibility == null) nextPossibility = new HashSet<>();
            for(Integer next : route) {
                nextPossibility.add(next);
                int[] tmp = {next.intValue(), here[1] + 1};
                que.add(tmp);
            }
            possibility2.put(Integer.valueOf(here[1] + 1), nextPossibility);
        }

        PrintWriter out =  new PrintWriter(System.out);
        for(int i = 0;i < 2*n;i++) {
            if(i != 0) {
                Set<Integer> set1 = possibility1.get(Integer.valueOf(i));
                Set<Integer> set2 = possibility2.get(Integer.valueOf(i - 1));
                Set<Integer> tmp = new HashSet<>(set1);
                tmp.removeAll(set2);
                if(tmp.isEmpty()) {
                    out.println(i - 1);
                    break;
                }
            }
            Set<Integer> tmp1 = possibility1.get(Integer.valueOf(i));
            Set<Integer> tmp2 = possibility2.get(Integer.valueOf(i));
            tmp1.removeAll(tmp2);
            if(tmp1.isEmpty()) {
                out.println(i);
                break;
            }
        }
        out.flush();
    }

    // 以下、ユーティリティ
    Set<long[]> primeFactorize(long n){
        // たとえnが10^18でも約数の個数は400を超えないため素因数はもっと少なくなる
        Set<long[]> ret = new HashSet<>(500);
        for(long p = 2;p * p <= n;p++) {
            if(n % p != 0) continue;
            long num = 0;
            while(n % p == 0) {
                ++num;
                n /= p;
            }
            long[] arr = {p, num};
            ret.add(arr);
        }
        if(n != 1) {
            long[] arr = {n, 1};
            ret.add(arr);
        }

        return ret;
    }

    // nの約数を求める
    Set<Long> searchDivisor(long n){
        Set<Long> ret = new HashSet<>(200);
        for(long i = 1;i * i <= n;i++) {
            if(n % i == 0) {
                ret.add(Long.valueOf(i));
                if(i * i != n) ret.add(Long.valueOf(n / i));
            }
        }

        return ret;
    }

    // 素数判定
    public static boolean isPrime(long num) {
        if (num < 2) return false;
        else if (num == 2) return true;
        else if (num % 2 == 0) return false;

        double sqrtNum = Math.sqrt(num);
        for (int i = 3; i <= sqrtNum; i += 2) {
            if (num % i == 0) return false;
        }

        return true;
    }

    long nCm(long n, long m) {
        if(n < m) return 0l;
        long c = 1l;
        m = (n - m < m ? n - m : m);
        for(long ns = n - m + 1, ms = 1;ms <= m;ns ++, ms++) {
            c *= ns;
            c /= ms;
        }
        return c;
    }

    long modnCm(long n, long m, long mod) {
        if(n < m) return 0l;
        long upper = 1l;
        long downer = 1l;
        m = (n - m < m ? n - m : m);
        for(long ns = n - m + 1, ms = 1;ms <= m;ns ++, ms++) {
            upper = upper * ns % mod;
            downer = downer * ms % mod;
        }
        return upper * modInv(downer, mod) % mod;
    }

    // mod m での a の逆元 を計算する
    long modInv(long a, long m) {
        long b = m, u = 1, v = 0;
        while(b > 0) {
            long t = a / b;
            a -= t * b;
            long tmp = a;
            a = b;
            b = tmp;
            u -= t * v;
            tmp = u;
            u = v;
            v = tmp;
        }
        u %= m;
        if (u < 0) u += m;
        return u;
    }

    long modPow(long a, long n, long mod) {
        long res = 1l;
        while(n > 0l) {
            if(n % 2l == 1l) res = res * a % mod;
            a = a * a % mod;
            n /= 2l;
        }
        return res;
    }

}

class UnionFind {
    protected int[] parent;

    UnionFind(int n) {
        parent = new int[n];
        for(int i = 0;i < n;i++) parent[i] = i;
    }

    public int rootOf(int x) {
        if(parent[x] == x) return x;
        else return parent[x] = rootOf(parent[x]);
    }

    public boolean same(int x, int y) {
        return rootOf(x) == rootOf(y);
    }

    public void unite(int x, int y) {
        x = rootOf(x);
        y = rootOf(y);
        if(x == y) return;
        parent[x] = y;
        return;
    }
}

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDouble() { return Double.parseDouble(next());}
}