結果

問題 No.1094 木登り / Climbing tree
ユーザー mkawa2mkawa2
提出日時 2021-06-10 16:38:07
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 929 ms / 2,000 ms
コード長 2,562 bytes
コンパイル時間 416 ms
コンパイル使用メモリ 82,568 KB
実行使用メモリ 140,680 KB
最終ジャッジ日時 2024-04-25 19:40:52
合計ジャッジ時間 21,905 ms
ジャッジサーバーID
(参考情報)
judge3 / judge5
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 37 ms
53,276 KB
testcase_01 AC 887 ms
130,420 KB
testcase_02 AC 233 ms
140,680 KB
testcase_03 AC 249 ms
80,544 KB
testcase_04 AC 355 ms
99,972 KB
testcase_05 AC 518 ms
122,112 KB
testcase_06 AC 455 ms
94,776 KB
testcase_07 AC 900 ms
129,228 KB
testcase_08 AC 903 ms
130,856 KB
testcase_09 AC 882 ms
129,524 KB
testcase_10 AC 874 ms
129,420 KB
testcase_11 AC 864 ms
129,496 KB
testcase_12 AC 909 ms
129,560 KB
testcase_13 AC 884 ms
129,904 KB
testcase_14 AC 881 ms
129,368 KB
testcase_15 AC 411 ms
91,468 KB
testcase_16 AC 588 ms
127,016 KB
testcase_17 AC 494 ms
106,968 KB
testcase_18 AC 460 ms
99,424 KB
testcase_19 AC 551 ms
118,396 KB
testcase_20 AC 897 ms
129,544 KB
testcase_21 AC 510 ms
108,484 KB
testcase_22 AC 929 ms
128,984 KB
testcase_23 AC 908 ms
130,496 KB
testcase_24 AC 883 ms
130,480 KB
testcase_25 AC 886 ms
130,060 KB
testcase_26 AC 889 ms
130,072 KB
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

# sys.setrecursionlimit(10**6)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LI1(): return list(map(int1, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
dij = [(0, 1), (1, 0), (1, 1), (1, -1)]
inf = 10**16
md = 998244353
# md = 10**9+7

class LCA:
    # 頂点は0~n-1
    def __init__(self, to, root=0):
        self.to = to
        self.n = len(to)
        self.parents = [-1]*(self.n+1)
        self.cost = [0]*self.n
        self.depth = [0]*self.n
        self.__dfs(root)
        self.max_level = max(self.depth).bit_length()
        self.ancestor = [self.parents]+[[-1]*(self.n+1) for _ in range(self.max_level)]
        row0 = self.ancestor[0]
        for lv in range(self.max_level):
            row1 = self.ancestor[lv+1]
            for u in range(self.n):
                row1[u] = row0[row0[u]]
            row0 = row1

    def __dfs(self, root):
        stack = [root]
        while stack:
            u = stack.pop()
            pu = self.parents[u]
            du = self.depth[u]
            for v, c in self.to[u]:
                if v == pu: continue
                self.parents[v] = u
                self.depth[v] = du+1
                self.cost[v] = self.cost[u]+c
                stack.append(v)

    # 最小共通祖先
    def anc(self, u, v):
        diff = self.depth[u]-self.depth[v]
        if diff < 0: u, v = v, u
        diff = abs(diff)
        lv = 0
        while diff:
            if diff & 1: u = self.ancestor[lv][u]
            lv, diff = lv+1, diff >> 1
        if u == v: return u
        for lv in range(self.depth[u].bit_length()-1, -1, -1):
            anclv = self.ancestor[lv]
            if anclv[u] != anclv[v]: u, v = anclv[u], anclv[v]
        return self.parents[u]

    def get_cost(self, u, v):
        w = self.anc(u, v)
        return self.cost[u]+self.cost[v]-2*self.cost[w]

n = II()
to = [[] for _ in range(n)]
for _ in range(n-1):
    u, v, c = LI1()
    to[u].append((v, c+1))
    to[v].append((u, c+1))
lca = LCA(to)
for _ in range(II()):
    u, v = LI1()
    print(lca.get_cost(u, v))
0