Fixed Length Paths II

Problem Overview

Attribute Value
Difficulty Hard
Category Tree Algorithms
Time Limit 1 second
Key Technique Centroid Decomposition + Fenwick Tree (BIT)
CSES Link https://cses.fi/problemset/task/2081

Learning Goals

After solving this problem, you will be able to:

  • Apply centroid decomposition to efficiently process tree path queries
  • Use Fenwick Tree (BIT) for range counting with efficient updates
  • Combine divide-and-conquer with data structures for O(n log^2 n) complexity

Problem Statement

Problem: Given a tree with n nodes, count the number of distinct paths with length in [k1, k2].

Input:

  • Line 1: Three integers n, k1, k2
  • Lines 2 to n: Two integers a, b (an edge)

Output:

  • Number of paths with length in range [k1, k2]

Constraints:

  • 1 <= n <= 2 * 10^5
  • 1 <= k1 <= k2 <= n - 1

Example

Input:
5 2 3
1 2
2 3
3 4
4 5

Output:
5

Explanation:

Tree: 1 --- 2 --- 3 --- 4 --- 5

Paths of length 2: (1,3), (2,4), (3,5) = 3 paths
Paths of length 3: (1,4), (2,5) = 2 paths
Total = 5

Intuition: How to Think About This Problem

Pattern Recognition

Key Question: How do we count paths of varying lengths without checking all O(n^2) pairs?

Centroid decomposition allows divide-and-conquer on trees. Every path either passes through the centroid or lies entirely within a subtree after removing it.

The Key Insight

  1. Find centroid (node whose removal creates subtrees of size <= n/2)
  2. Count all paths passing through this centroid using BIT for range queries
  3. Remove centroid and recursively solve for each subtree

For paths in range [k1, k2]: when processing a node at depth d, the partner must be at depth in [k1-d, k2-d]. BIT enables O(log n) range queries.

Solution 1: Brute Force

Idea

For each pair of nodes, compute distance using BFS and count if in [k1, k2].

Code

def solve_brute_force(n, k1, k2, edges):
  """
  Time: O(n^2), Space: O(n)
  """
  from collections import defaultdict, deque

  adj = defaultdict(list)
  for a, b in edges:
    adj[a].append(b)
    adj[b].append(a)

  count = 0
  for start in range(1, n + 1):
    dist = [-1] * (n + 1)
    dist[start] = 0
    queue = deque([start])

    while queue:
      u = queue.popleft()
      for v in adj[u]:
        if dist[v] == -1:
          dist[v] = dist[u] + 1
          queue.append(v)
          if k1 <= dist[v] <= k2:
            count += 1

  return count // 2  # Each path counted twice

Complexity

Metric Value Explanation
Time O(n^2) BFS from each node
Space O(n) Distance array and queue

Solution 2: Optimal (Centroid Decomposition + BIT)

Key Insight

The Trick: Use centroid decomposition to count each path exactly once when it passes through some centroid. Use BIT for O(log n) range queries.

Dry Run Example

Input: n=5, k1=2, k2=3, tree: 1-2-3-4-5

Step 1: Find centroid = 3

Step 2: Process centroid 3
  Left subtree (via 2): depths = [1, 2]  (nodes 2, 1)
  Right subtree (via 4): depths = [1, 2] (nodes 4, 5)

  Add left depths to BIT, then query for right:
    d=1: query [1,2] -> 2 matches (pairs with depth 1,2)
    d=2: query [0,1] -> 1 match (pair with depth 1)
  Paths through centroid: 3

Step 3: Recurse - remaining subtrees have no paths in [2,3]

Total = 5 paths: (1,3), (2,4), (3,5) at length 2; (1,4), (2,5) at length 3

Code (Python)

import sys
from collections import defaultdict
sys.setrecursionlimit(300000)

def solve(n, k1, k2, edges):
  adj = [[] for _ in range(n + 1)]
  for a, b in edges:
    adj[a].append(b)
    adj[b].append(a)

  bit = [0] * (n + 2)

  def bit_update(i, delta):
    i += 1
    while i <= n + 1:
      bit[i] += delta
      i += i & (-i)

  def bit_query(i):
    if i < 0: return 0
    i = min(i + 1, n + 1)
    s = 0
    while i > 0:
      s += bit[i]
      i -= i & (-i)
    return s

  def bit_range(lo, hi):
    return 0 if lo > hi else bit_query(hi) - bit_query(lo - 1)

  removed = [False] * (n + 1)
  size = [0] * (n + 1)

  def get_size(u, p):
    size[u] = 1
    for v in adj[u]:
      if v != p and not removed[v]:
        get_size(v, u)
        size[u] += size[v]

  def get_centroid(u, p, tree_size):
    for v in adj[u]:
      if v != p and not removed[v] and size[v] > tree_size // 2:
        return get_centroid(v, u, tree_size)
    return u

  def get_depths(u, p, d, depths):
    if d > k2: return
    depths.append(d)
    for v in adj[u]:
      if v != p and not removed[v]:
        get_depths(v, u, d + 1, depths)

  result = 0

  def process(u):
    nonlocal result
    get_size(u, -1)
    c = get_centroid(u, -1, size[u])
    removed[c] = True

    max_d = 0
    for v in adj[c]:
      if removed[v]: continue
      depths = []
      get_depths(v, c, 1, depths)

      for d in depths:
        lo, hi = max(0, k1 - d), k2 - d
        if hi >= 0:
          result += bit_range(lo, hi)

      for d in depths:
        bit_update(d, 1)
        max_d = max(max_d, d)

    for d in range(max_d + 1):
      v = bit_range(d, d)
      if v > 0: bit_update(d, -v)

    for v in adj[c]:
      if not removed[v]:
        process(v)

  process(1)
  return result

def main():
  data = sys.stdin.read().split()
  n, k1, k2 = int(data[0]), int(data[1]), int(data[2])
  edges = [(int(data[3+2*i]), int(data[4+2*i])) for i in range(n-1)]
  print(solve(n, k1, k2, edges))

if __name__ == "__main__":
  main()

Complexity

Metric Value Explanation
Time O(n log^2 n) O(log n) decomposition levels, O(log n) BIT ops per node
Space O(n) Adjacency list, BIT, recursion stack

Common Mistakes

Mistake 1: Incorrect BIT Cleanup

# WRONG - O(n) per centroid makes total O(n^2)
bit = [0] * (n + 2)

Fix: Only clear depths actually used.

Mistake 2: Counting Intra-Subtree Paths

# WRONG - counts paths within same subtree
all_depths = []
for v in adj[centroid]:
  get_depths(v, centroid, 1, all_depths)
# Then counting pairs in all_depths

Fix: Process subtrees sequentially, query BIT before adding new depths.

Mistake 3: Missing Depth 0

Problem: Paths from node at depth d to centroid (depth 0) not counted. Fix: Initialize BIT[0]=1 or handle separately when k1 <= d <= k2.

Edge Cases

Case Input Expected Why
Single edge n=2, k1=1, k2=1 1 One path
Star graph n=5, k1=2, k2=2 6 C(4,2)=6 through center
k1 = k2 n=5, k1=2, k2=2 varies Exact length only
Large k1 n=5, k1=10, k2=10 0 No path >= n

When to Use This Pattern

Use Centroid Decomposition + BIT When:

  • Counting/summing over all paths in a tree
  • Path length constraints involve ranges
  • Need better than O(n^2) complexity

Pattern Recognition Checklist:

  • Counting paths in tree? Consider centroid decomposition
  • Range constraints on length? Add BIT/segment tree
  • Combining subtree info? Process subtrees sequentially

Easier (Do First)

Problem Why It Helps
Fixed Length Paths I (CSES 2080) Exact length k, simpler
Tree Diameter (CSES 1131) Basic tree traversal

Similar Difficulty

Problem Key Difference
Path Queries II (CSES 2134) Different operations

Harder (Do After)

Problem New Concept
Distance Queries (CSES 1135) LCA + preprocessing

Key Takeaways

  1. Core Idea: Centroid decomposition ensures every path passes through exactly one centroid
  2. Time Optimization: O(n^2) to O(n log^2 n) via centroid decomposition + BIT
  3. Pattern: Canonical template for “path counting in range” problems

Additional Resources