Count the unique pairs of tree vertices that share an edge length of exactly k within a tree structure
In the realm of computer science, a common problem arises when we need to determine if a path exists in a tree with K vertices, where at most D of them are at a distance apart. This article focuses on a solution to a related problem: counting the number of distinct pairs of vertices in a tree with n nodes that have a distance of exactly k.
The solution to this problem can be achieved using dynamic programming (DP), a technique that breaks down a complex problem into smaller, more manageable sub-problems. This approach is based on depth-first search (DFS) and distance counting arrays.
### Key Idea:
- Perform a DFS from any node (often the root). - At each node, maintain a frequency array `dp` where `dp[d]` = number of nodes in the subtree of the current node at distance `d` from it. - When combining results from child subtrees, count how many pairs between different subtrees have the distance exactly `k`. - Use a bottom-up DP approach to accumulate the count.
### Step-by-step approach:
1. **Initialization:** - Pick any node as root. - For each node `u`, maintain an array `dp[u][0..k]` where `dp[u][d]` = number of nodes in subtree rooted at `u` that are at distance `d` from `u`.
2. **DFS and DP computation:** - For a leaf node, `dp[u] = 1` (the node itself) and `dp[u][d] = 0` for `d > 0`. - For an internal node `u`, after computing `dp` arrays for all its children, combine them to calculate: - The number of pairs `(a, b)` where `a` is in one child's subtree and `b` is in another child's subtree such that the distance between `a` and `b` is exactly `k`. - This count is accumulated by iterating over pairs of distances `(d1, d2)` from different child subtrees where `d1 + d2 + 2 = k` (distance between nodes through their lowest common ancestor `u`). - Update `dp[u]` by shifting the distances from children by 1, i.e., \[ dp[u][d] = \sum_{\text{child } v} dp[v][d-1] \] for `d > 0`, and `dp[u] = 1`.
3. **Combine counts from subtrees:** - While merging children `c1` and `c2` of node `u`, count pairs with \[ \sum_{d1=0}^{k-2} dp[c1][d1] \times dp[c2][k - 2 - d1] \] - Sum this over all pairs of distinct children.
4. **Answer:** - The total count obtained after processing the root is the number of pairs at distance exactly `k`.
### Complexity: - This approach runs in **O(N * k)** time where N is the number of nodes, because you do a DFS and at each node you process its children arrays up to length `k`.
### Intuition: - The DP arrays allow efficient counting of how many nodes are at each distance in subtrees. - Pairs with distance `k` crossing subtrees can be counted by combining the dp arrays from children. - By doing this bottom-up, you avoid expensive recomputations.
An example algorithm outline in pseudocode is provided below:
```python def dfs(u, parent): dp[u] = [0] * (k+1) dp[u][0] = 1 total_pairs = 0
for v in adj[u]: if v == parent: continue pairs_from_child = dfs(v, u) total_pairs += pairs_from_child
# Count pairs between dp[u] and dp[v] for d1 in range(k-1): d2 = k - 2 - d1 if d2 >= 0: total_pairs += dp[u][d1] * dp[v][d2]
# Merge dp[v] into dp[u], shifting distances by 1 for d in range(k-1, 0, -1): dp[u][d] += dp[v][d-1]
return total_pairs ```
Run DFS from root: `answer = dfs(root, -1)`
This approach is a standard tree-DP form for distance pair counting problems and closely aligns with techniques outlined in similar problems on platforms like GeeksforGeeks or competitive programming tutorials.
For a detailed explanation and example code, see approaches discussed on GeeksforGeeks for counting pairs at a certain distance in trees. The problem examples provided show that when k = 2, the answer is 4, and when k = 3, the answer is 2.
- The dynamic programming approach used in this solution can be likened to data-and-cloud-computing technology, where a complex problem is broken down into smaller sub-problems to be efficiently solved using clever algorithms.
- The counting of distinct pairs in a tree using a trie-like structure (depth-first search, distance counting arrays) is an example of how modern technology, in this case, dynamic programming, can solve complex computer science problems.