結果

問題 No.898 tri-βutree
ユーザー tktk_snsn
提出日時 2020-06-05 19:33:57
言語 Python3
(3.13.1 + numpy 2.2.1 + scipy 1.14.1)
結果
AC  
実行時間 3,754 ms / 4,000 ms
コード長 1,724 bytes
コンパイル時間 249 ms
コンパイル使用メモリ 13,184 KB
実行使用メモリ 121,284 KB
最終ジャッジ日時 2024-12-16 07:03:47
合計ジャッジ時間 60,305 ms
ジャッジサーバーID
(参考情報)
judge4 / judge3
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 * 21
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 #

from collections import defaultdict
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)

N = int(input())
D = N.bit_length()
edge = [[] for _ in range(N)]
for _ in range(N - 1):
    a, b, c = map(int, input().split())
    edge[a].append((b, c))
    edge[b].append((a, c))
Q = int(input())
query = [sorted(list(map(int, input().split()))) for _ in range(Q)]


root = 0
dist = [-1] * N
dist[root] = 0
depth = [0] * N
parent = [[-1] * N for _ in range(D)]

node = [root]
while node:
    s = node.pop()
    d = dist[s]
    dep = depth[s]
    for t, c in edge[s]:
        if dist[t] != -1:
            continue
        dist[t] = d + c
        depth[t] = dep + 1
        node.append(t)
        parent[0][t] = s

for i in range(D - 1):
    for j in range(N):
        parent[i + 1][j] = parent[i][parent[i][j]]


def lower_ancester(x, h):
    for i in reversed(range(D)):
        if h >= (1 << i):
            h -= (1 << i)
            x = parent[i][x]
    return x


def LCA(u, v):
    if depth[u] < depth[v]:
        u, v = v, u
    u = lower_ancester(u, depth[u] - depth[v])
    if u == v:
        return u
    for i in reversed(range(D)):
        if parent[i][u] != parent[i][v]:
            u = parent[i][u]
            v = parent[i][v]
    return parent[0][u]


path = defaultdict(int)
for x, y, z in query:
    if not path[(x, y)]:
        lca = LCA(x, y)
        path[(x, y)] = dist[x] + dist[y] - 2 * dist[lca]
    if not path[(y, z)]:
        lca = LCA(y, z)
        path[(y, z)] = dist[y] + dist[z] - 2 * dist[lca]
    if not path[(x, z)]:
        lca = LCA(x, z)
        path[(x, z)] = dist[x] + dist[z] - 2 * dist[lca]
    size = (path[(x, y)] + path[(y, z)] + path[(x, z)]) // 2
    print(size)
0