結果

問題 No.898 tri-βutree
ユーザー mkawa2mkawa2
提出日時 2021-03-24 12:15:00
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 2,742 bytes
コンパイル時間 329 ms
コンパイル使用メモリ 11,244 KB
実行使用メモリ 59,728 KB
最終ジャッジ日時 2023-08-17 22:02:43
合計ジャッジ時間 26,333 ms
ジャッジサーバーID
（参考情報） 		judge15 / judge14
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 970 ms
59,728 KB
testcase_01 AC 16 ms
8,620 KB
testcase_02 WA -
testcase_03 WA -
testcase_04 WA -
testcase_05 WA -
testcase_06 WA -
testcase_07 WA -
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
testcase_20 WA -
testcase_21 WA -
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)]
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.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, _ in self.to[u]:
                if v == pu: continue
                self.parents[v] = u
                self.depth[v] = du+1
                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]

n = II()
to = [[] for _ in range(n)]
for _ in range(n-1):
    u, v, w = LI()
    to[u].append((v, w))
    to[v].append((u, w))

lca = LCA(to)
stack = [(0, -1)]
cost = [0]*n
while stack:
    u, pu = stack.pop()
    for v, w in to[u]:
        if v == pu: continue
        cost[v] = cost[u]+w
        stack.append((v, u))

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

for _ in range(II()):
    ans = 0
    x, y, z = LI()
    if lca.depth[x] > lca.depth[y]: x, y = y, x
    if lca.depth[x] > lca.depth[z]: x, z = z, x
    p, c = cal(y, z)
    ans += c
    _, c = cal(p, x)
    ans += c
    print(ans)
0