3326. Count Pairs of Connectable Servers in a Weighted Tree Network

Medium
Array
Tree
Depth-First Search

Description

You are given an unrooted weighted tree with n vertices representing servers numbered from 0 to n - 1, an array edges where edges[i] = [ai, bi, weighti] represents a bidirectional edge between vertices ai and bi of weight weighti. You are also given an integer signalSpeed.

Two servers a and b are connectable through a server c if:

  • a < b, a != c and b != c.
  • The distance from c to a is divisible by signalSpeed.
  • The distance from c to b is divisible by signalSpeed.
  • The path from c to b and the path from c to a do not share any edges.

Return an integer array count of length n where count[i] is the number of server pairs that are connectable through the server i.

 

Example 1:

Input: edges = [[0,1,1],[1,2,5],[2,3,13],[3,4,9],[4,5,2]], signalSpeed = 1
Output: [0,4,6,6,4,0]
Explanation: Since signalSpeed is 1, count[c] is equal to the number of pairs of paths that start at c and do not share any edges.
In the case of the given path graph, count[c] is equal to the number of servers to the left of c multiplied by the servers to the right of c.

Example 2:

Input: edges = [[0,6,3],[6,5,3],[0,3,1],[3,2,7],[3,1,6],[3,4,2]], signalSpeed = 3
Output: [2,0,0,0,0,0,2]
Explanation: Through server 0, there are 2 pairs of connectable servers: (4, 5) and (4, 6).
Through server 6, there are 2 pairs of connectable servers: (4, 5) and (0, 5).
It can be shown that no two servers are connectable through servers other than 0 and 6.

 

Constraints:

  • 2 <= n <= 1000
  • edges.length == n - 1
  • edges[i].length == 3
  • 0 <= ai, bi < n
  • edges[i] = [ai, bi, weighti]
  • 1 <= weighti <= 106
  • 1 <= signalSpeed <= 106
  • The input is generated such that edges represents a valid tree.

Hints

Hint 1
Take each node as the root of the tree, run DFS, and save for each node <code>i</code>, the number of nodes in the subtree rooted at <code>i</code> whose distance to the root is divisible by <code>signalSpeed</code>.
Hint 2
If the root has <code>m</code> children named <code>c<sub>1</sub>, c<sub>2</sub>, …, c<sub>m</sub></code> that respectively have <code>num[c<sub>1</sub>], num[c<sub>2</sub>], …, num[c<sub>m</sub>]</code> nodes in their subtrees whose distance is divisible by signalSpeed. Then, there are <code>((S - num[c<sub>i</sub>]) * num[c<sub>i</sub>]) / 2</code>that are connectable through the root that we have fixed, where <code>S</code> is the sum of <code>num[c<sub>i</sub>]</code>.

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