結果

問題 No.2337 Equidistant
コンテスト
ユーザー gew1fw
提出日時 2025-06-12 17:58:29
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,710 bytes
コンパイル時間 173 ms
コンパイル使用メモリ 82,472 KB
実行使用メモリ 575,496 KB
最終ジャッジ日時 2025-06-12 17:59:29
合計ジャッジ時間 27,842 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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 * 5 WA * 22 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
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
    
    MAX_LOG = 20
    parent = [[-1]*(N+1) for _ in range(MAX_LOG)]
    depth = [0]*(N+1)
    in_time = [0]*(N+1)
    out_time = [0]*(N+1)
    sz = [0]*(N+1)
    time = 0
    
    from sys import setrecursionlimit
    setrecursionlimit(1 << 25)
    def dfs(u, p):
        nonlocal time
        in_time[u] = time
        time +=1
        parent[0][u] = p
        depth[u] = depth[p] +1 if p != -1 else 0
        sz[u] =1
        for v in edges[u]:
            if v != p:
                dfs(v, u)
                sz[u] += sz[v]
        out_time[u] = time
        time +=1
    dfs(1, -1)
    
    for k in range(1, MAX_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 lca(u, v):
        if depth[u] < depth[v]:
            u, v = v, u
        for k in range(MAX_LOG-1, -1, -1):
            if depth[u] - (1 << k) >= depth[v]:
                u = parent[k][u]
        if u == v:
            return u
        for k in range(MAX_LOG-1, -1, -1):
            if parent[k][u] != parent[k][v]:
                u = parent[k][u]
                v = parent[k][v]
        return parent[0][u]
    
    def get_kth_ancestor(u, k):
        if k <0:
            return -1
        for i in range(MAX_LOG-1, -1, -1):
            if k >= (1 << i):
                u = parent[i][u]
                k -= (1 << i)
                if u == -1:
                    return -1
        return u if k ==0 else -1
    
    def is_ancestor(u, v):
        return in_time[u] <= in_time[v] and out_time[v] <= out_time[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)
        else:
            k_prime = k - a
            steps = b - k_prime
            M = get_kth_ancestor(T, steps)
        
        u = None
        if is_ancestor(M, S):
            delta = depth[S] - depth[M] -1
            if delta <0:
                u_candidate = S
                if parent[0][u_candidate] == M:
                    u = u_candidate
                else:
                    u = None
            else:
                X = get_kth_ancestor(S, delta)
                if X != -1 and parent[0][X] == M:
                    u = X
                else:
                    u = None
        else:
            u = None
        
        v = None
        if is_ancestor(M, T):
            delta = depth[T] - depth[M] -1
            if delta <0:
                v_candidate = T
                if parent[0][v_candidate] == M:
                    v = v_candidate
                else:
                    v = None
            else:
                X = get_kth_ancestor(T, delta)
                if X != -1 and parent[0][X] == M:
                    v = X
                else:
                    v = None
        else:
            v = None
        
        ans = sz[M]
        if u is not None:
            ans -= sz[u]
        if v is not None:
            ans -= sz[v]
        print(ans)

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