結果

問題 No.2337 Equidistant
ユーザー gew1fw
提出日時 2025-06-12 17:58:19
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,262 bytes
コンパイル時間 157 ms
コンパイル使用メモリ 82,520 KB
実行使用メモリ 539,416 KB
最終ジャッジ日時 2025-06-12 17:59:17
合計ジャッジ時間 24,871 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 1 WA * 26 MLE * 1
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 sys import stdin
from collections import deque

sys.setrecursionlimit(1 << 25)

def main():
    input = sys.stdin.read
    data = input().split()
    idx = 0
    N, Q = int(data[idx]), int(data[idx+1])
    idx +=2

    edges = [[] for _ in range(N+1)]
    for _ in range(N-1):
        a = int(data[idx])
        b = int(data[idx+1])
        edges[a].append(b)
        edges[b].append(a)
        idx +=2

    LOG = 20
    parent = [[-1]*(N+1) for _ in range(LOG)]
    depth = [0]*(N+1)
    size = [1]*(N+1)

    # BFS to setup parent[0], depth, and size
    q = deque([1])
    parent[0][1] = -1
    visited = [False]*(N+1)
    visited[1] = True
    while q:
        u = q.popleft()
        for v in edges[u]:
            if not visited[v] and v != parent[0][u]:
                parent[0][v] = u
                depth[v] = depth[u] +1
                visited[v] = True
                q.append(v)

    # DFS to compute size
    def dfs(u, p):
        for v in edges[u]:
            if v != p:
                dfs(v, u)
                size[u] += size[v]
    dfs(1, -1)

    # Binary lifting for LCA and k-th ancestor
    for k in range(1, LOG):
        for v in range(1, N+1):
            if parent[k-1][v] != -1:
                parent[k][v] = parent[k-1][parent[k-1][v]]
            else:
                parent[k][v] = -1

    def get_kth_ancestor(u, k):
        for i in range(LOG):
            if k & (1 << i):
                u = parent[i][u]
                if u == -1:
                    return -1
        return u

    def lca(u, v):
        if depth[u] < depth[v]:
            u, v = v, u
        for k in reversed(range(LOG)):
            if depth[u] - (1 << k) >= depth[v]:
                u = parent[k][u]
        if u == v:
            return u
        for k in reversed(range(LOG)):
            if parent[k][u] != -1 and parent[k][u] != parent[k][v]:
                u = parent[k][u]
                v = parent[k][v]
        return parent[0][u]

    for _ in range(Q):
        S = int(data[idx])
        T = int(data[idx+1])
        idx +=2

        L = lca(S, T)
        a = depth[S] - depth[L]
        b = depth[T] - depth[L]
        D = a + b
        if D % 2 != 0:
            print(0)
            continue

        k = D // 2
        if k <= a:
            M = get_kth_ancestor(S, k)
            u = get_kth_ancestor(S, k-1) if k > 0 else -1
            if M == L:
                v = get_kth_ancestor(T, b-1) if b > 0 else -1
            else:
                steps = depth[L] - depth[M] -1
                if steps <0:
                    v = -1
                else:
                    v = get_kth_ancestor(L, steps)
        else:
            m_val = k - a
            M = get_kth_ancestor(T, m_val)
            u = get_kth_ancestor(T, m_val-1) if m_val >0 else -1
            if M == L:
                v = get_kth_ancestor(S, a-1) if a >0 else -1
            else:
                steps = depth[L] - depth[M] -1
                if steps <0:
                    v = -1
                else:
                    v = get_kth_ancestor(L, steps)

        if u == -1 or v == -1:
            print(0)
            continue

        answer = size[M] - size[u] - size[v]
        print(answer)

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