목차

회사 문화 4

ps
링크acmicpc.net/…
출처BOJ
문제 번호14288
문제명회사 문화 4
레벨플래티넘 3
분류

구간 쿼리

시간복잡도O(n+mlogn)
인풋사이즈n<=100,000, m<=100,000
사용한 언어Python
제출기록64496KB / 764ms
최고기록764ms
해결날짜2021/04/30

풀이

코드

"""Solution code for "BOJ 14288. 회사 문화 4".

- Problem link: https://www.acmicpc.net/problem/14288
- Solution link: http://www.teferi.net/ps/problems/boj/14288
"""

import sys
from teflib import fenwicktree
from teflib import tgraph


def euler_tour_technique(tree, root):
    subtree_ranges = [[None, None] for _ in tree]
    order = 0
    for node in tgraph.dfs(tree, root, yields_on_leave=True):
        if subtree_ranges[node][0] is None:
            subtree_ranges[node][0] = order
            order += 1
        else:
            subtree_ranges[node][1] = order
    return subtree_ranges


def main():
    n, m = [int(x) for x in sys.stdin.readline().split()]
    tree = [[] for _ in range(n)]
    bosses = [int(x) for x in sys.stdin.readline().split()]
    for employee, boss in enumerate(bosses):
        if boss != -1:
            tree[boss - 1].append(employee)
    subtree_ranges = euler_tour_technique(tree, 0)
    fenwick_for_upward = fenwicktree.FenwickTree(n + 1)
    fenwick_for_downward = fenwicktree.FenwickTree(n + 1)
    is_downward = True
    for _ in range(m):
        query = [int(x) for x in sys.stdin.readline().split()]
        if query[0] == 1:
            _, i, w = query
            beg, end = subtree_ranges[i - 1]
            if is_downward:
                fenwick_for_downward.update(beg, w)
                fenwick_for_downward.update(end, -w)
            else:
                fenwick_for_upward.update(beg, w)
        elif query[0] == 2:
            i = query[1]
            beg, end = subtree_ranges[i - 1]
            print(fenwick_for_downward.query(0, beg + 1)
                  + fenwick_for_upward.query(beg, end))
        else:
            is_downward = not is_downward


if __name__ == '__main__':
    main()