結果

問題 No.2337 Equidistant
コンテスト
ユーザー lam6er
提出日時 2025-04-16 00:59:45
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 3,061 bytes
コンパイル時間 146 ms
コンパイル使用メモリ 82,912 KB
実行使用メモリ 198,888 KB
最終ジャッジ日時 2025-04-16 01:01:15
合計ジャッジ時間 22,293 ms
ジャッジサーバーID
（参考情報） 		judge1 / 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 WA * 20 TLE * 1 -- * 7
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().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
    
    LOG = 20
    parent = [[0]*(LOG+1) for _ in range(N+1)]
    depth = [0]*(N+1)
    sz = [1]*(N+1)
    visited = [False]*(N+1)
    
    from collections import deque
    q = deque([1])
    visited[1] = True
    parent[1][0] = 0
    
    while q:
        u = q.popleft()
        for v in adj[u]:
            if not visited[v] and v != parent[u][0]:
                parent[v][0] = u
                depth[v] = depth[u] + 1
                visited[v] = True
                q.append(v)
    
    stack = [(1, False)]
    while stack:
        u, visited_flag = stack.pop()
        if visited_flag:
            for v in adj[u]:
                if parent[v][0] == u:
                    sz[u] += sz[v]
            continue
        stack.append((u, True))
        for v in adj[u]:
            if parent[v][0] == u and not visited_flag:
                stack.append((v, False))
    
    for k in range(1, LOG+1):
        for u in range(1, N+1):
            parent[u][k] = parent[parent[u][k-1]][k-1]
    
    def lca(u, v):
        if depth[u] < depth[v]:
            u, v = v, u
        for k in range(LOG, -1, -1):
            if depth[u] - (1 << k) >= depth[v]:
                u = parent[u][k]
        if u == v:
            return u
        for k in range(LOG, -1, -1):
            if parent[u][k] != parent[v][k]:
                u = parent[u][k]
                v = parent[v][k]
        return parent[u][0]
    
    def get_kth_ancestor(u, k):
        current = u
        for i in range(LOG, -1, -1):
            if k >= (1 << i):
                current = parent[current][i]
                k -= (1 << i)
        return current if k == 0 else 0
    
    def find_child(u, mid):
        current = u
        while current != 0 and parent[current][0] != mid:
            current = parent[current][0]
        return current if parent[current][0] == mid else None
    
    for _ in range(Q):
        S = int(input[idx])
        T = int(input[idx+1])
        idx += 2
        
        if S == T:
            print(0)
            continue
        
        L = lca(S, T)
        a = depth[S] - depth[L]
        b = depth[T] - depth[L]
        D = a + b
        
        if D % 2 != 0:
            print(0)
            continue
        
        mid_dist = D // 2
        if mid_dist <= a:
            mid = get_kth_ancestor(S, mid_dist)
        else:
            mid = get_kth_ancestor(T, b - (mid_dist - a))
        
        u = find_child(S, mid)
        v = find_child(T, mid)
        
        if u is None or v is None:
            print(0)
            continue
        
        ans = sz[mid] - sz[u] - sz[v]
        print(ans if ans >= 0 else 0)

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