結果

問題 No.2337 Equidistant
ユーザー lam6er
提出日時 2025-03-26 15:53:38
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,043 bytes
コンパイル時間 283 ms
コンパイル使用メモリ 82,288 KB
実行使用メモリ 272,212 KB
最終ジャッジ日時 2025-03-26 15:54:50
合計ジャッジ時間 35,248 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
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 * 6 WA * 22
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 #
プレゼンテーションモードにする

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
    depth = [0] * (N + 1)
    parent = [0] * (N + 1)
    q = deque([1])
    depth[1] = 0
    parent[1] = 0
    while q:
        u = q.popleft()
        for v in adj[u]:
            if v != parent[u] and depth[v] == 0:
                depth[v] = depth[u] + 1
                parent[v] = u
                q.append(v)
    LOG = 20
    up = [[0] * (N + 1) for _ in range(LOG)]
    up[0] = parent
    for k in range(1, LOG):
        for v in range(1, N + 1):
            up[k][v] = up[k-1][up[k-1][v]]
    size = [1] * (N + 1)
    stack = [(1, False)]
    while stack:
        u, visited = stack.pop()
        if visited:
            for v in adj[u]:
                if v != parent[u]:
                    size[u] += size[v]
            continue
        stack.append((u, True))
        for v in adj[u]:
            if v != parent[u]:
                stack.append((v, False))
    edge_map = {}
    for u in range(1, N+1):
        for v in adj[u]:
            if depth[u] < depth[v]:
                s = size[v]
                edge_map[(u, v)] = s
                edge_map[(v, u)] = N - s
            else:
                s = size[u]
                edge_map[(v, u)] = s
                edge_map[(u, v)] = N - s
    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 = up[k][u]
        if u == v:
            return u
        for k in reversed(range(LOG)):
            if up[k][u] != up[k][v]:
                u = up[k][u]
                v = up[k][v]
        return parent[u]
    def get_kth(u, k):
        current = u
        for i in range(LOG):
            if k & (1 << i):
                current = up[i][current]
                if current == 0:
                    return 0
        return current
    output = []
    for _ in range(Q):
        S = int(input[idx])
        T = int(input[idx+1])
        idx += 2
        L = lca(S, T)
        dS = depth[S] - depth[L]
        dT = depth[T] - depth[L]
        D = dS + dT
        if D % 2 != 0:
            output.append('0')
            continue
        k = D // 2
        if k <= dS:
            M = get_kth(S, k)
        else:
            M = get_kth(T, (dS + dT) - k)
        if k == 0:
            u = S
        else:
            u = get_kth(S, k - 1)
        steps_t = (D - k) - 1
        if steps_t < 0:
            v = T
        else:
            v = get_kth(T, steps_t)
        a = edge_map.get((M, u), 0)
        b = edge_map.get((M, v), 0)
        ans = N - a - b
        output.append(str(ans))
    print('\n'.join(output))
if __name__ == '__main__':
    main()
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0