結果

問題 No.898 tri-βutree
ユーザー OKCH3COOHOKCH3COOH
提出日時 2019-10-26 22:11:36
言語 Python3
(3.12.2 + numpy 1.26.4 + scipy 1.12.0)
結果
WA  
実行時間 -
コード長 2,362 bytes
コンパイル時間 536 ms
コンパイル使用メモリ 12,800 KB
実行使用メモリ 71,552 KB
最終ジャッジ日時 2024-09-14 19:04:43
合計ジャッジ時間 14,533 ms
ジャッジサーバーID
(参考情報)
judge3 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2,870 ms
68,352 KB
testcase_01 AC 31 ms
10,880 KB
testcase_02 WA -
testcase_03 WA -
testcase_04 WA -
testcase_05 WA -
testcase_06 WA -
testcase_07 TLE -
testcase_08 TLE -
testcase_09 TLE -
testcase_10 TLE -
testcase_11 TLE -
testcase_12 TLE -
testcase_13 TLE -
testcase_14 TLE -
testcase_15 TLE -
testcase_16 TLE -
testcase_17 TLE -
testcase_18 TLE -
testcase_19 TLE -
testcase_20 TLE -
testcase_21 TLE -
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 deque

class LCA:
    def __init__(self, size):
        self.size = size
        self.level = (size - 1).bit_length()
        self.edges = [[] for _ in range(size)]
        self.depth = [0] * size
        self.length = [float('inf')] * size
        self.prev = [None] * size
        self.kprev = None

    def addEdge(self, fr, to, cost=1):
        self.edges[fr].append((to, cost))
        self.edges[to].append((fr, cost))

    def dfs(self):
        st = deque([(0, 0)])
        self.depth[0] = 0
        self.length[0] = 0
        while st:
            now, prev = st.pop()
            for to, cost in self.edges[now]:
                if to == prev:
                    continue
                st.append((to, now))
                self.depth[to] = self.depth[now] + 1
                self.length[to] = self.length[now] + cost
                self.prev[to] = now

    def construct(self):
        self.dfs()
        kprev = [self.prev]
        S = self.prev
        for _ in range(self.level):
            T = [0] * self.size
            for i in range(self.size):
                if S[i] is None:
                    continue
                T[i] = S[S[i]]
            kprev.append(T)
            S = T
        self.kprev = kprev

    def lca(self, u, v):
        dist = self.depth[v] - self.depth[u]
        if dist < 0:
            u, v = v, u
            dist = -dist

        for k in range(self.level + 1):
            if dist & 1:
                v = self.kprev[k][v]
            dist //= 2

        if u == v:
            return u

        for k in range(self.level)[:: -1]:
            prevU = self.kprev[k][u]
            prevV = self.kprev[k][v]
            if prevU != prevV:
                u = prevU
                v = prevV

        return self.kprev[0][u]

    def dist(self, u, v):
        return self.length[u] + self.length[v] - self.length[self.lca(u, v)] * 2

N = int(input())

tree = LCA(N)

for _ in range(N - 1):
    fr, to, cost = map(int, input().split())
    tree.addEdge(fr, to, cost)

tree.construct()
Q = int(input())
ans = []
for _ in range(Q):
    x, y, z = map(int, input().split())
    ans.append(min(
        tree.dist(x, y) + tree.dist(tree.lca(x, z), z),
        tree.dist(y, z) + tree.dist(tree.lca(z, y), x),
        tree.dist(z, x) + tree.dist(tree.lca(y, x), y),
    ))

print(*ans, sep='\n')
0