結果

問題 No.1778 括弧列クエリ / Bracketed Sequence Query
コンテスト
ユーザー norioc
提出日時 2025-03-26 00:14:13
言語 Python3
(3.13.1 + numpy 2.2.1 + scipy 1.14.1)
結果
TLE  
実行時間 -
コード長 4,194 bytes
コンパイル時間 640 ms
コンパイル使用メモリ 12,928 KB
実行使用メモリ 91,352 KB
最終ジャッジ日時 2025-03-26 00:14:32
合計ジャッジ時間 18,753 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
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ファイルパターン 結果
sample -- * 1
other AC * 8 TLE * 1 -- * 18
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
from itertools import count
import sys
sys.setrecursionlimit(10 ** 6)
#input = sys.stdin.readline

try:
    import pypyjit
    pypyjit.set_param('max_unroll_recursion=-1')
except ModuleNotFoundError:
    pass


class LCA:
    def __init__(self, n, adj, root=0):
        K = 1
        while (1 << K) < n: K += 1
        parent = [[-1] * n for _ in range(K)]
        dist = [-1] * n

        def dfs():
            s = [(root, -1, 0)]
            while s:
                v, par, depth = s.pop()
                parent[0][v] = par
                dist[v] = depth
                for to in adj[v]:
                    if to == par: continue
                    s.append((to, v, depth+1))

        dfs()
        for k in range(K-1):
            for v in range(n):
                if parent[k][v] < 0: continue
                parent[k+1][v] = parent[k][parent[k][v]]
        self.parent = parent
        self.dist = dist

    def query(self, u, v) -> int:
        """二頂点 u, v の LCA"""
        if self.dist[u] < self.dist[v]: u, v = v, u
        K = len(self.parent)
        # LCA までの距離を同じにする
        for k in range(K):
            if (self.dist[u] - self.dist[v]) >> k & 1:
                u = self.parent[k][u]
        # 二分探索で LCA を求める
        if u == v: return u
        for k in range(K-1, -1, -1):
            if self.parent[k][u] != self.parent[k][v]:
                u = self.parent[k][u]
                v = self.parent[k][v]
        return self.parent[0][u]

    def distance(self, u, v) -> int:
        """二頂点 u, v の距離"""
        return self.dist[u] + self.dist[v] - 2 * self.dist[self.query(u, v)]

    def is_on_path(self, u, v, a) -> bool:
        """二頂点 u, v 上に a があるか"""
        return self.distance(u, a) + self.distance(a, v) == self.distance(u, v)


class Reader:
    def __init__(self, data: list):
        self.data = data
        self.p = 0

    def has_next(self) -> bool:
        return self.p < len(self.data)

    def peek(self):
        assert self.has_next()

        return self.data[self.p]

    def next_indexed(self):
        assert self.has_next()

        v = self.data[self.p]
        p = self.p
        self.p += 1
        return v, p

    def next(self):
        assert self.has_next()

        v = self.data[self.p]
        self.p += 1
        return v


class Node:
    def __init__(self, left, right, children):
        self.left = left    # 開き括弧のインデックス
        self.right = right  # 閉じ括弧のインデックス
        self.children = children

    def tree_size(self) -> int:
        res = 1
        for child in self.children:
            res += child.tree_size()

        return res


def make_tree(reader: Reader) -> Node:
    assert reader.peek() == '('

    v, l = reader.next_indexed()
    children = []
    while reader.has_next():
        match reader.peek():
            case ')':
                _, r = reader.next_indexed()
                return Node(l, r, children)

            case '(':
                c = make_tree(reader)
                children.append(c)

    assert False


def traverse(node: Node):

    def dfs(node, node_id, adj, ind2id, id2node, counter):
        id2node[node_id] = node
        ind2id[node.left] = node_id
        ind2id[node.right] = node_id

        for child in node.children:
            id = next(counter)
            adj[node_id].append(id)
            adj[id].append(node_id)


            dfs(child, id, adj, ind2id, id2node, counter)

    counter = count()
    root_id = next(counter)
    adj = defaultdict(list)
    ind2id = {}
    id2node = {}
    dfs(node, root_id, adj, ind2id, id2node, counter)

    return adj, ind2id, id2node


N, Q = map(int, input().split())
S = input()

s = '(' + S + ')'
reader = Reader(list(s))
root = make_tree(reader)
adj, ind2id, id2node = traverse(root)

lca = LCA(len(id2node), adj, root=0)

for _ in range(Q):
    x, y = map(int, input().split())

    a = ind2id[x]
    b = ind2id[y]
    root = lca.query(a, b)
    if root == 0:
        print(-1)
    else:
        node = id2node[root]
        print(f'{node.left} {node.right}')
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