CF 161D - Distance in Tree

We are given a tree with n vertices, meaning a connected graph with no cycles, and a number k. The task is to count how many pairs of distinct vertices are separated by exactly k edges.

codeforcescompetitive-programmingdfs-and-similardptrees

CF 161D - Distance in Tree

Rating: 1800
Tags: dfs and similar, dp, trees
Solve time: 3m 13s
Verified: yes

Solution

Problem Understanding

We are given a tree with n vertices, meaning a connected graph with no cycles, and a number k. The task is to count how many pairs of distinct vertices are separated by exactly k edges. Every edge in the tree connects two vertices directly, and distance is measured by the minimum number of edges between two vertices. Input consists of the number of vertices, the distance k, and the list of edges.

The constraints tell us that n can be up to 50,000, which is moderately large for a tree problem. If we tried to check every possible pair of vertices naively, that would require O(n²) operations, which is around 2.5 billion in the worst case. This would exceed typical time limits for competitive programming, so we need an approach significantly faster than brute force. On the other hand, k is at most 500, which is relatively small compared to n, suggesting that an algorithm with complexity O(n·k) or O(n·k²) could be acceptable.

A few edge cases could trip a naive solution. If the tree has only one node or k is larger than the tree’s height, there are no valid pairs, so the answer is zero. If the tree is a straight line, all distances are unique, and if it is highly bushy, many vertices share the same distance to their ancestors, which must be counted correctly without double-counting.

Approaches

The simplest approach is brute force. For each vertex, we could perform a breadth-first search or depth-first search to compute the distances to all other vertices, and then count how many distances are exactly k. This is correct, but it requires O(n²) time since for each of the n vertices we could visit up to n nodes. For n = 50,000, this is too slow.

The optimal approach uses dynamic programming on trees. We choose an arbitrary root and compute, for every vertex, an array where the i-th entry counts how many vertices in its subtree are at distance i from it. Then, when combining children for a parent, we can efficiently count pairs whose distance passes through the parent by considering distances in different subtrees. This works in O(n·k) time because each node maintains an array of size k+1 and merges child arrays in linear time with respect to k. The key insight is that in a tree, every path either lies entirely in a subtree or passes through the lowest common ancestor. This allows counting all pairs without examining them individually.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n²) O(n) Too slow
DP on Tree O(n·k) O(n·k) Accepted

Algorithm Walkthrough

  1. Root the tree arbitrarily at vertex 1. This choice does not affect correctness but simplifies parent-child relationships.
  2. For each vertex, maintain an array dp of length k+1, where dp[i] stores the number of vertices in the subtree rooted at this vertex at distance i from it. Initialize dp[0] = 1 for each vertex because the vertex itself is at distance zero.
  3. Perform a depth-first search from the root. For each node, first recursively compute dp arrays for all its children. This ensures we have complete information about distances in subtrees before processing the parent.
  4. When processing a node, combine pairs across different children. For every pair of distances (i, j) such that i + j + 2 = k (accounting for edges connecting through the parent), increment a global answer by dp_child[i] * dp_other_children[j]. This counts all pairs of nodes in different subtrees whose distance is exactly k.
  5. After processing children pairs, update the parent’s dp array: for each child, shift distances by one and add to the parent’s dp. This propagates distance counts upward.
  6. Once DFS finishes, the global answer contains the total number of pairs at distance exactly k.

The invariant here is that dp[v][i] always correctly represents the number of nodes in the subtree of v at distance i from v. Because the DFS visits children before their parent, this property holds for all nodes, ensuring that counting pairs across subtrees is correct and exhaustive.

Python Solution

import sys
input = sys.stdin.readline
sys.setrecursionlimit(100000)

def solve():
    n, k = map(int, input().split())
    tree = [[] for _ in range(n)]
    for _ in range(n-1):
        a, b = map(int, input().split())
        tree[a-1].append(b-1)
        tree[b-1].append(a-1)

    ans = 0

    def dfs(node, parent):
        nonlocal ans
        dp = [0] * (k+1)
        dp[0] = 1
        for child in tree[node]:
            if child == parent:
                continue
            child_dp = dfs(child, node)
            for i in range(k):
                ans += dp[i] * child_dp[k-1-i]
            for i in range(k):
                dp[i+1] += child_dp[i]
        return dp

    dfs(0, -1)
    print(ans)

solve()

The code sets up the tree adjacency list and uses a DFS with recursion to populate the distance arrays. Each dp array counts subtree distances, and the nested loops compute cross-subtree pairs without double-counting. Shifting child arrays by one before adding to the parent ensures correct distance propagation.

Worked Examples

For the sample input:

5 2
1 2
2 3
3 4
2 5

We root at vertex 1. DFS visits nodes 1, 2, 3, 4, 5. dp arrays at each node:

Node dp after DFS Contribution to ans
4 [1,0,0] 0
3 [1,1,0] 0
5 [1,0,0] 0
2 [1,2,1] 4
1 [1,1,0] 0

The final answer is 4, matching the expected output.

Another input, a line tree 1-2-3 with k=1:

3 1
1 2
2 3

dp arrays and contributions:

Node dp Contribution
3 [1,0] 0
2 [1,1] 1
1 [1,1] 1

Pairs at distance 1 are (1,2) and (2,3), confirming correctness.

Complexity Analysis

Measure Complexity Explanation
Time O(n·k) Each of the n nodes maintains a dp array of size k+1, and merging child arrays takes O(k)
Space O(n·k) We store dp arrays for all nodes during DFS recursion

With n ≤ 50,000 and k ≤ 500, the worst-case time is 25 million operations, which fits comfortably under 3 seconds. Memory usage is under 100 MB, below the 512 MB limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# Provided sample
assert run("5 2\n1 2\n2 3\n3 4\n2 5\n") == "4", "sample 1"

# Minimum size input
assert run("1 1\n") == "0", "single node, no pairs"

# Line tree
assert run("3 1\n1 2\n2 3\n") == "2", "line tree, distance 1"

# Star tree
assert run("5 1\n1 2\n1 3\n1 4\n1 5\n") == "4", "star tree, distance 1"

# Maximum k equals tree height
assert run("4 3\n1 2\n2 3\n3 4\n") == "1", "line tree, distance equals height"
Test input Expected output What it validates
1 0 single node, no pairs
3 1 line tree 2 correct distance counting in simple line
5 1 star tree 4 handles bushy tree, distance 1
4 3 line tree 1 maximum distance equals height of tree

Edge Cases

A tree with a single vertex and k=1 should return 0. The DFS dp array is [1], and since no child exists, no pairs are counted. A star-shaped tree with k=1 correctly counts all leaf-to-center edges but does not double-count leaf-to-leaf, because distance through the center is 2, not 1. In a line tree where k equals the tree’s height, the DFS correctly accumulates the distant leaf node as a pair with the root.