結果

問題 No.2337 Equidistant
コンテスト
ユーザー gew1fw
提出日時 2025-06-12 17:58:15
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,263 bytes
コンパイル時間 455 ms
コンパイル使用メモリ 82,256 KB
実行使用メモリ 229,900 KB
最終ジャッジ日時 2025-06-12 17:58:53
合計ジャッジ時間 31,660 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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 * 4 WA * 24
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 bisect import bisect_left
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)
    in_time = [0]*(N+1)
    out_time = [0]*(N+1)
    size = [1]*(N+1)
    children = [[] for _ in range(N+1)]
    time = 0

    stack = [(1, -1, False)]
    while stack:
        u, p, visited = stack.pop()
        if visited:
            out_time[u] = time
            for v in edges[u]:
                if v != p:
                    size[u] += size[v]
            continue
        in_time[u] = time
        time +=1
        parent[0][u] = p
        stack.append((u, p, True))
        for v in edges[u]:
            if v != p:
                depth[v] = depth[u] +1
                stack.append((v, u, False))
                children[u].append(v)

    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]]

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

    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

    for u in range(1, N+1):
        children[u].sort(key=lambda x: in_time[x])

    def find_child(u, x_in):
        child_list = children[u]
        left, right = 0, len(child_list)
        while left < right:
            mid = (left + right) //2
            v = child_list[mid]
            if in_time[v] > x_in:
                right = mid
            else:
                left = mid +1
        left -=1
        if left >=0:
            v = child_list[left]
            if in_time[v] <= x_in <= out_time[v]:
                return v
        p = parent[0][u]
        if p != -1 and in_time[u] <= x_in <= out_time[u]:
            return p
        return None

    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)
        else:
            M = get_kth_ancestor(T, (d -k))
        u = find_child(M, in_time[S])
        v = find_child(M, in_time[T])
        ans = size[M]
        if u is not None:
            ans -= size[u]
        if v is not None:
            ans -= size[v]
        print(ans)

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