結果

問題 No.363 門松サイクル
ユーザー maspy
提出日時 2020-05-16 16:12:07
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 764 ms / 4,000 ms
コード長 2,443 bytes
コンパイル時間 268 ms
コンパイル使用メモリ 82,444 KB
実行使用メモリ 140,276 KB
最終ジャッジ日時 2024-09-22 11:49:10
合計ジャッジ時間 13,360 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge4
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ファイルパターン 結果
sample AC * 2
other AC * 27
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

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

N = int(readline())
A = (0, ) + tuple(map(int, readline().split()))
G = [[] for _ in range(N + 1)]
for _ in range(N - 1):
    x, y = map(int, readline().split())
    G[x].append(y)
    G[y].append(x)
Q = int(readline())
m = map(int, read().split())
query = tuple(zip(m, m))

parent = [0] * (N + 1)
depth = [0] * (N + 1)
stack = []
order = []
root = 1
stack = [root]
while stack:
    v = stack.pop()
    order.append(v)
    for w in G[v]:
        if w == parent[v]:
            continue
        parent[w] = v
        depth[w] = depth[v] + 1
        stack.append(w)

K = 18
P = [parent]
for _ in range(K):
    p = P[-1]
    P.append([p[v] for v in p])

def LCA(u, v):
    du, dv = depth[u], depth[v]
    if du < dv:
        du, dv = dv, du
        u, v = v, u
    n = du - dv
    for i in range(K):
        if n & 1:
            u = P[i][u]
        n >>= 1
    if u == v:
        return u
    for i in range(K - 1, -1, -1):
        u1, v1 = P[i][u], P[i][v]
        if u1 == v1:
            continue
        u, v = u1, v1
    return parent[u]

def is_kadomatsu(a, b, c):
    if a == c:
        return False
    return (a > b < c) or (a < b > c)

longest_kadomatsu = [0] * (N + 1)
for v in order[1:]:
    p = parent[v]
    pp = parent[p]
    if not p:
        longest_kadomatsu[v] = 0
    elif not pp:
        longest_kadomatsu[v] = 1
    else:
        a, b, c = A[v], A[p], A[pp]
        if is_kadomatsu(a, b, c):
            longest_kadomatsu[v] = longest_kadomatsu[p] + 1
        else:
            longest_kadomatsu[v] = 1

def ascend(v, k):
    for i in range(K):
        if k & 1:
            v = P[i][v]
        k >>= 1
    return v

longest_kadomatsu

def solve(u,v):
    w = LCA(u,v)
    if v == w:
        u,v = v,u
    du, dv, dw = depth[u], depth[v], depth[w]
    if longest_kadomatsu[v] < dv - dw:
        return False
    if longest_kadomatsu[u] < du - dw:
        return False
    if u == w:
        v1 = parent[v]
        v2 = ascend(v, dv-dw-1)
        return is_kadomatsu(A[v2], A[w], A[v]) and is_kadomatsu(A[w], A[v], A[v1])
    v1 = parent[v]
    v2 = ascend(v, dv-dw-1)
    u1 = parent[u]
    u2 = ascend(u, du-dw-1)
    return is_kadomatsu(A[v2], A[w], A[u2]) and is_kadomatsu(A[v], A[u], A[u1]) and is_kadomatsu(A[u], A[v], A[v1])        

query

for a,b in query:
    print('YES' if solve(a,b) else 'NO')
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