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3067. Count Pairs of Connectable Servers in a Weighted Tree Network

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.

Solutions

Solution 1: Enumeration + DFS

First, we construct an adjacency list g based on the edges given in the problem, where g[a] represents all the neighbor nodes of node a and their corresponding edge weights.

Then, we can enumerate each node a as the connecting intermediate node, and calculate the number of nodes t that start from the neighbor node b of a and whose distance to node a can be divided by signalSpeed through depth-first search. Then, the number of connectable node pairs of node a increases by s * t, where s represents the cumulative number of nodes that start from the neighbor node b of a and whose distance to node a cannot be divided by signalSpeed. Then we update s to s + t.

After enumerating all nodes a, we can get the number of connectable node pairs for all nodes.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$, where $n$ is the number of nodes.

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class Solution:
    def countPairsOfConnectableServers(
        self, edges: List[List[int]], signalSpeed: int
    ) -> List[int]:
        def dfs(a: int, fa: int, ws: int) -> int:
            cnt = 0 if ws % signalSpeed else 1
            for b, w in g[a]:
                if b != fa:
                    cnt += dfs(b, a, ws + w)
            return cnt

        n = len(edges) + 1
        g = [[] for _ in range(n)]
        for a, b, w in edges:
            g[a].append((b, w))
            g[b].append((a, w))
        ans = [0] * n
        for a in range(n):
            s = 0
            for b, w in g[a]:
                t = dfs(b, a, w)
                ans[a] += s * t
                s += t
        return ans

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class Solution {
    private int signalSpeed;
    private List<int[]>[] g;

    public int[] countPairsOfConnectableServers(int[][] edges, int signalSpeed) {
        int n = edges.length + 1;
        g = new List[n];
        this.signalSpeed = signalSpeed;
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1], w = e[2];
            g[a].add(new int[] {b, w});
            g[b].add(new int[] {a, w});
        }
        int[] ans = new int[n];
        for (int a = 0; a < n; ++a) {
            int s = 0;
            for (var e : g[a]) {
                int b = e[0], w = e[1];
                int t = dfs(b, a, w);
                ans[a] += s * t;
                s += t;
            }
        }
        return ans;
    }

    private int dfs(int a, int fa, int ws) {
        int cnt = ws % signalSpeed == 0 ? 1 : 0;