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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using ii = pair<int, int>;

const int MAX { 200010 };
const ll MOD { 1000000007 }, oo { 1LL << 62 };

vector<int> adj[MAX];
struct LCA
{
    vector<int> height, euler, first, segtree;
    vector<bool> visited;
    int n;

    //LCA(vector<vector<int>> &adj, int root = 0) {
    LCA(int N, int root = 0) {
        n = N;
        height.resize(n);
        first.resize(n);
        euler.reserve(n * 2);
        visited.assign(n, false);
        dfs(root);
        int m = euler.size();
        segtree.resize(m * 4);
        build(1, 0, m - 1);
    }

    void dfs(int node, int h = 0) {
        visited[node] = true;
        height[node] = h;
        first[node] = euler.size();
        euler.push_back(node);
        for (auto to : adj[node]) {
            if (!visited[to]) {
                dfs(to, h + 1);
                euler.push_back(node);
            }
        }
    }

    void build(int node, int b, int e) {
        if (b == e) {
            segtree[node] = euler[b];
        } else {
            int mid = (b + e) / 2;
            build(node << 1, b, mid);
            build(node << 1 | 1, mid + 1, e);
            int l = segtree[node << 1], r = segtree[node << 1 | 1];
            segtree[node] = (height[l] < height[r]) ? l : r;
        }
    }

    int query(int node, int b, int e, int L, int R) {
        if (b > R || e < L)
            return -1;
        if (b >= L && e <= R)
            return segtree[node];
        int mid = (b + e) >> 1;

        int left = query(node << 1, b, mid, L, R);
        int right = query(node << 1 | 1, mid + 1, e, L, R);
        if (left == -1) return right;
        if (right == -1) return left;
        return height[left] < height[right] ? left : right;
    }

    int lca(int u, int v) {
        int left = first[u], right = first[v];
        if (left > right)
            swap(left, right);
        return query(1, 0, euler.size() - 1, left, right);
    }
};

int ps[MAX], is_leaf[MAX];

void dfs(int u, int p)
{
    ps[u] = p;

    bool leaf = true;

    for (auto v : adj[u])
    {
        if (v != p)
        {
            leaf = false;
            dfs(v, u);
        }
    }

    is_leaf[u] = leaf;
}

vector<int> path(int u, int v)
{
#ifdef DEBUG
    cout << "Calculando path de " << u << " a " << v << endl;
#endif
    dfs(u, 0);

    vector<int> vs(1, v);

    int p = ps[v];

    while (p)
    {
        vs.push_back(p);
        p = ps[p];
    }

    reverse(vs.begin(), vs.end());

    return vs;
}

ll solve(int N, int a, int b)
{
    LCA xs(N + 1, a);

#ifdef DEBUG
    for (int u = 1; u <= N; ++u)
        cout << "u = " << u << ", dist = " << xs.height[u] << endl;
#endif
    int v = 0, d = 0, g = 0;

    for (int u = 1; u <= N; ++u)
    {
        if (u == a or u == b)
            continue;

        auto p = xs.lca(u, b);

#ifdef DEBUG
cout << "u = " << u << ", b = " << b << ", p = " << p << endl;
#endif

        if (is_leaf[u] and xs.height[p] < (xs.height[b] - xs.height[p]) and xs.height[u] > d)
        {
            v = u;
            d = xs.height[u];
            g = p;
#ifdef DEBUG
cout << "--- v = " << u << ", d = " << d << endl;
#endif
        }
    }

    if (v == 0)
    {
        v = adj[a].front();
#ifdef DEBUG
cout << "!!! não há um melhor caminho: pegando um filho qualquer: v = " << v << endl;
#endif
    }

    vector<int> p = path(a, v), q = path(b, g ? g : a);

#ifdef DEBUG
    cout << "p = ";
    for (auto x : p)
        cout << x << ' ';
    cout << endl;
    cout << "q = ";
    for (auto x : q)
        cout << x << ' ';
    cout << endl;
#endif
    int ans = 0, x = 0, y = 0;
    N = p.size();
    int M = q.size();
    
    while (y < M - 1)
    {
        if (x < N - 1)
            ++x;
        else
            x = N - 2;
#ifdef DEBUG
    cout << "Takahashi foi para " << p[x] << endl;
#endif
        if (p[x] == q[y])
            return ans;

        ++ans;
        ++y;

#ifdef DEBUG
    cout << "Aoki foi para " << q[y] << ", ans = " << ans << endl;
#endif

        if (q[y] == p[x])
            return ans;
    }

    q = path(q.back(), v);

#ifdef DEBUG
    cout << "new q = ";
    for (auto x : q)
        cout << x << ' ';
    cout << endl;
#endif

    M = q.size();
    y = 0;

    while (y < M - 1)
    {
        if (x < N - 1)
            ++x;
        else
            x = N - 2;
#ifdef DEBUG
    cout << "Takahashi foi para " << p[x] << endl;
#endif
        if (p[x] == q[y])
            return ans;

        ++ans;
        ++y;

#ifdef DEBUG
    cout << "Aoki foi para " << q[y] << ", ans = " << ans << endl;
#endif

        if (q[y] == p[x])
            return ans;
    }

    return ans;
}

int main()
{
    ios::sync_with_stdio(false);

    int N, u, v;
    cin >> N >> u >> v;

    vector<ll> xs(N);

    for (int i = 0; i < N - 1; ++i)
    {
        int a, b;
        cin >> a >> b;

        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    auto ans = solve(N, u, v);

    cout << ans << '\n';

    return 0;
}