結果

問題 No.2337 Equidistant
コンテスト
ユーザー lam6er
提出日時 2025-04-15 21:50:43
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,296 ms / 4,000 ms
コード長 2,884 bytes
コンパイル時間 363 ms
コンパイル使用メモリ 82,140 KB
実行使用メモリ 204,956 KB
最終ジャッジ日時 2025-04-15 21:52:47
合計ジャッジ時間 19,496 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 1
other AC * 28
権限があれば一括ダウンロードができます

ソースコード

diff # 

import sys
from collections import deque

sys.setrecursionlimit(1 << 25)

def main():
    input = sys.stdin.read().split()
    idx = 0
    N, Q = int(input[idx]), int(input[idx+1])
    idx += 2
    
    adj = [[] for _ in range(N+1)]
    for _ in range(N-1):
        a = int(input[idx])
        b = int(input[idx+1])
        adj[a].append(b)
        adj[b].append(a)
        idx += 2
    
    parent = [0] * (N + 1)
    depth = [0] * (N + 1)
    size = [1] * (N + 1)
    visited = [False] * (N + 1)
    q = deque([1])
    visited[1] = True
    
    while q:
        u = q.popleft()
        for v in adj[u]:
            if not visited[v] and v != parent[u]:
                visited[v] = True
                parent[v] = u
                depth[v] = depth[u] + 1
                q.append(v)
    
    order = []
    visited = [False] * (N + 1)
    q = deque([1])
    visited[1] = True
    while q:
        u = q.popleft()
        order.append(u)
        for v in adj[u]:
            if not visited[v] and v != parent[u]:
                visited[v] = True
                q.append(v)
    
    for u in reversed(order):
        for v in adj[u]:
            if v != parent[u]:
                size[u] += size[v]
    
    log_level = 20
    up = [[0] * (N + 1) for _ in range(log_level)]
    up[0] = parent[:]
    for k in range(1, log_level):
        for u in range(1, N + 1):
            up[k][u] = up[k-1][up[k-1][u]]
    
    def lca(u, v):
        if depth[u] < depth[v]:
            u, v = v, u
        for k in range(log_level-1, -1, -1):
            if depth[u] - (1 << k) >= depth[v]:
                u = up[k][u]
        if u == v:
            return u
        for k in range(log_level-1, -1, -1):
            if up[k][u] != up[k][v]:
                u = up[k][u]
                v = up[k][v]
        return up[0][u]
    
    def get_kth_ancestor(u, k):
        current = u
        for i in range(log_level-1, -1, -1):
            if k >= (1 << i):
                current = up[i][current]
                k -= (1 << i)
        return current if k == 0 else -1
    
    output = []
    for _ in range(Q):
        S = int(input[idx])
        T = int(input[idx+1])
        idx += 2
        
        ancestor = lca(S, T)
        a = depth[S] - depth[ancestor]
        b = depth[T] - depth[ancestor]
        D = a + b
        
        if D % 2 != 0:
            output.append(0)
            continue
        
        k = D // 2
        if k <= a:
            M = get_kth_ancestor(S, k)
        else:
            M = get_kth_ancestor(T, k)
        
        u = get_kth_ancestor(S, k-1)
        v = get_kth_ancestor(T, k-1)
        
        size_u = size[u] if parent[u] == M else N - size[M]
        size_v = size[v] if parent[v] == M else N - size[M]
        
        output.append(N - size_u - size_v)
    
    print('\n'.join(map(str, output)))

if __name__ == '__main__':
    main()
0