{ "data": { "question": { "questionId": "3863", "questionFrontendId": "3607", "categoryTitle": "Algorithms", "boundTopicId": 3714205, "title": "Power Grid Maintenance", "titleSlug": "power-grid-maintenance", "content": "
You are given an integer c
representing c
power stations, each with a unique identifier id
from 1 to c
(1‑based indexing).
These stations are interconnected via n
bidirectional cables, represented by a 2D array connections
, where each element connections[i] = [ui, vi]
indicates a connection between station ui
and station vi
. Stations that are directly or indirectly connected form a power grid.
Initially, all stations are online (operational).
\n\nYou are also given a 2D array queries
, where each query is one of the following two types:
[1, x]
: A maintenance check is requested for station x
. If station x
is online, it resolves the check by itself. If station x
is offline, the check is resolved by the operational station with the smallest id
in the same power grid as x
. If no operational station exists in that grid, return -1.
[2, x]
: Station x
goes offline (i.e., it becomes non-operational).
Return an array of integers representing the results of each query of type [1, x]
in the order they appear.
Note: The power grid preserves its structure; an offline (non‑operational) node remains part of its grid and taking it offline does not alter connectivity.
\n\n\n
Example 1:
\n\nInput: c = 5, connections = [[1,2],[2,3],[3,4],[4,5]], queries = [[1,3],[2,1],[1,1],[2,2],[1,2]]
\n\nOutput: [3,2,3]
\n\nExplanation:
\n\n{1, 2, 3, 4, 5}
are online and form a single power grid.[1,3]
: Station 3 is online, so the maintenance check is resolved by station 3.[2,1]
: Station 1 goes offline. The remaining online stations are {2, 3, 4, 5}
.[1,1]
: Station 1 is offline, so the check is resolved by the operational station with the smallest id
among {2, 3, 4, 5}
, which is station 2.[2,2]
: Station 2 goes offline. The remaining online stations are {3, 4, 5}
.[1,2]
: Station 2 is offline, so the check is resolved by the operational station with the smallest id
among {3, 4, 5}
, which is station 3.Example 2:
\n\nInput: c = 3, connections = [], queries = [[1,1],[2,1],[1,1]]
\n\nOutput: [1,-1]
\n\nExplanation:
\n\n[1,1]
: Station 1 is online in its isolated grid, so the maintenance check is resolved by station 1.[2,1]
: Station 1 goes offline.[1,1]
: Station 1 is offline and there are no other stations in its grid, so the result is -1.\n
Constraints:
\n\n1 <= c <= 105
0 <= n == connections.length <= min(105, c * (c - 1) / 2)
connections[i].length == 2
1 <= ui, vi <= c
ui != vi
1 <= queries.length <= 2 * 105
queries[i].length == 2
queries[i][0]
is either 1 or 2.1 <= queries[i][1] <= c
给你一个整数 c
,表示 c
个电站,每个电站有一个唯一标识符 id
,从 1 到 c
编号。
这些电站通过 n
条 双向 电缆互相连接,表示为一个二维数组 connections
,其中每个元素 connections[i] = [ui, vi]
表示电站 ui
和电站 vi
之间的连接。直接或间接连接的电站组成了一个 电网 。
最初,所有 电站均处于在线(正常运行)状态。
\n\n另给你一个二维数组 queries
,其中每个查询属于以下 两种类型之一 :
[1, x]
:请求对电站 x
进行维护检查。如果电站 x
在线,则它自行解决检查。如果电站 x
已离线,则检查由与 x
同一 电网 中 编号最小 的在线电站解决。如果该电网中 不存在 任何 在线 电站,则返回 -1。
[2, x]
:电站 x
离线(即变为非运行状态)。
返回一个整数数组,表示按照查询中出现的顺序,所有类型为 [1, x]
的查询结果。
注意:电网的结构是固定的;离线(非运行)的节点仍然属于其所在的电网,且离线操作不会改变电网的连接性。
\n\n\n\n
示例 1:
\n\n输入: c = 5, connections = [[1,2],[2,3],[3,4],[4,5]], queries = [[1,3],[2,1],[1,1],[2,2],[1,2]]
\n\n输出: [3,2,3]
\n\n解释:
\n\n{1, 2, 3, 4, 5}
都在线,并组成一个电网。[1,3]
:电站 3 在线,因此维护检查由电站 3 自行解决。[2,1]
:电站 1 离线。剩余在线电站为 {2, 3, 4, 5}
。[1,1]
:电站 1 离线,因此检查由电网中编号最小的在线电站解决,即电站 2。[2,2]
:电站 2 离线。剩余在线电站为 {3, 4, 5}
。[1,2]
:电站 2 离线,因此检查由电网中编号最小的在线电站解决,即电站 3。示例 2:
\n\n输入: c = 3, connections = [], queries = [[1,1],[2,1],[1,1]]
\n\n输出: [1,-1]
\n\n解释:
\n\n[1,1]
:电站 1 在线,且属于其独立电网,因此维护检查由电站 1 自行解决。[2,1]
:电站 1 离线。[1,1]
:电站 1 离线,且其电网中没有其他电站,因此结果为 -1。\n\n
提示:
\n\n1 <= c <= 105
0 <= n == connections.length <= min(105, c * (c - 1) / 2)
connections[i].length == 2
1 <= ui, vi <= c
ui != vi
1 <= queries.length <= 2 * 105
queries[i].length == 2
queries[i][0]
为 1 或 2。1 <= queries[i][1] <= c
[2, x]
, remove x
from the set of its component",
"For query [1, x]
, if x
is in its component’s set return x
; otherwise if the set is non-empty return its smallest element; else return -1
",
"Precompute all components and then handle each query in O(log n) time using the sorted sets"
],
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