Question

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

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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.

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

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;
    for (var e : g[a]) {
      int b = e[0], w = e[1];
      if (b != fa) {
        cnt += dfs(b, a, ws + w);
      }
    }
    return cnt;
  }
}

class Solution {
public:
  vector<int> countPairsOfConnectableServers(vector<vector<int>>& edges, int signalSpeed) {
    int n = edges.size() + 1;
    vector<pair<int, int>> g[n];
    for (auto& e : edges) {
      int a = e[0], b = e[1], w = e[2];
      g[a].emplace_back(b, w);
      g[b].emplace_back(a, w);
    }
    function<int(int, int, int)> dfs = [&](int a, int fa, int ws) {
      int cnt = ws % signalSpeed == 0;
      for (auto& [b, w] : g[a]) {
        if (b != fa) {
          cnt += dfs(b, a, ws + w);
        }
      }
      return cnt;
    };
    vector<int> ans(n);
    for (int a = 0; a < n; ++a) {
      int s = 0;
      for (auto& [b, w] : g[a]) {
        int t = dfs(b, a, w);
        ans[a] += s * t;
        s += t;
      }
    }
    return ans;
  }
};

func countPairsOfConnectableServers(edges [][]int, signalSpeed int) []int {
  n := len(edges) + 1
  type pair struct{ x, w int }
  g := make([][]pair, n)
  for _, e := range edges {
    a, b, w := e[0], e[1], e[2]
    g[a] = append(g[a], pair{b, w})
    g[b] = append(g[b], pair{a, w})
  }
  var dfs func(a, fa, ws int) int
  dfs = func(a, fa, ws int) int {
    cnt := 0
    if ws%signalSpeed == 0 {
      cnt++
    }
    for _, e := range g[a] {
      b, w := e.x, e.w
      if b != fa {
        cnt += dfs(b, a, ws+w)
      }
    }
    return cnt
  }
  ans := make([]int, n)
  for a := 0; a < n; a++ {
    s := 0
    for _, e := range g[a] {
      b, w := e.x, e.w
      t := dfs(b, a, w)
      ans[a] += s * t
      s += t
    }
  }
  return ans
}

function countPairsOfConnectableServers(edges: number[][], signalSpeed: number): number[] {
  const n = edges.length + 1;
  const g: [number, number][][] = Array.from({ length: n }, () => []);
  for (const [a, b, w] of edges) {
    g[a].push([b, w]);
    g[b].push([a, w]);
  }
  const dfs = (a: number, fa: number, ws: number): number => {
    let cnt = ws % signalSpeed === 0 ? 1 : 0;
    for (const [b, w] of g[a]) {
      if (b != fa) {
        cnt += dfs(b, a, ws + w);
      }
    }
    return cnt;
  };
  const ans: number[] = Array(n).fill(0);
  for (let a = 0; a < n; ++a) {
    let s = 0;
    for (const [b, w] of g[a]) {
      const t = dfs(b, a, w);
      ans[a] += s * t;
      s += t;
    }
  }
  return ans;
}