CF 102697006 - Distance

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Solve time: 49s
Verified: yes

Solution

Problem Understanding

The problem asks for the number of unordered pairs of vertices in a tree whose shortest path contains exactly k edges. The input describes an undirected tree with n vertices, followed by the target distance k. The output is a single integer representing how many different pairs of vertices satisfy this distance requirement.

A tree has a unique path between every pair of vertices, which is the property that makes the problem tractable. We do not need to search for shortest paths repeatedly because there is only one possible path between any two nodes.

The important constraint is that n can reach 50,000 while k is at most 500. An algorithm that checks every pair of vertices would need about n² / 2 distance computations, which is around 1.25 billion operations at the largest size. That is too slow in a normal contest environment. The small value of k suggests that we should avoid considering long distances and focus only on paths of length up to k.

Several edge cases can break straightforward implementations.

For a single vertex tree, there are no pairs to count.

Input:

1 1

The correct output is:

A solution that assumes at least one edge exists may fail while building adjacency lists or initializing recursion.

When k = 1, every valid pair is simply an edge of the tree. For example:

3 1
1 2
2 3

The correct output is:

A method designed only for pairs that share a deeper common ancestor may miss these direct connections.

A chain can also expose double counting mistakes. For:

the correct output is:

The valid pairs are (1,3), (2,4), and (3,5). Counting paths from both endpoints independently without a way to avoid duplicates would count every pair twice.

Approaches

A direct solution would try every pair of vertices. For each pair, we could run a breadth-first search or use the tree structure to compute their distance, then increment the answer if the distance equals k. This is correct because every possible pair is examined.

The problem is the cost. There are roughly n² pairs, and even a simple constant-time distance calculation would still be too much when n = 50000. With BFS from every vertex, the complexity becomes O(n²), or worse depending on the implementation details, which is far beyond the limit.

The key observation is that every valid path of length k has a unique middle point or middle edge. If we choose a centroid of the tree, every path either lies completely inside one remaining component after removing that centroid, or it passes through the centroid. This lets us count all paths passing through the centroid, remove the smaller subproblems, and solve them recursively.

For a centroid, we only need to know how many vertices exist at each distance from it. If a vertex in one subtree is distance d from the centroid, then a vertex in another subtree at distance k-d creates a valid pair. We combine information from different subtrees while making sure we never count two vertices from the same subtree at the same centroid step.

The brute-force approach works because every pair is checked directly, but fails because there are too many pairs. The centroid decomposition observation reduces the problem into smaller trees, allowing every vertex to participate in only logarithmically many decomposition levels.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(n)	Too slow
Centroid Decomposition	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

Find the centroid of the current tree component. A centroid is a vertex such that after removing it, every remaining connected component has at most half of the original vertices. This guarantees that recursive decomposition has logarithmic depth.
Count all valid paths that pass through this centroid. Start with a frequency array where distance 0 contains the centroid itself. For each child subtree, collect how many vertices exist at every distance from the centroid. Before adding that subtree's distances into the global frequency array, query how many already processed vertices are at distance k - current_distance. Those pairs are valid because they come from different subtrees and their path goes through the centroid.
Add the current subtree's distances into the frequency array. The order matters because vertices inside the same subtree should not be paired during this centroid processing step. Their paths do not pass through the centroid and will be handled later when that subtree becomes a separate recursive component.
Mark the centroid as removed and recursively solve every remaining component. Each component is independent because every path inside it cannot use the removed centroid.
Continue until all components are processed. The sum of all counts found during centroid processing is the final answer.

Why it works:

Every pair of vertices has exactly one highest-level centroid in the decomposition where the two vertices become separated into different components or one of them is the centroid itself. At that moment, their path passes through that centroid and is counted exactly once. If both vertices remain in the same component, the pair is not counted yet, but it will be considered at a deeper decomposition level. Since every possible pair is handled at its unique separating centroid, no pair is missed or duplicated.

Python Solution

import sys
input = sys.stdin.readline

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

    removed = [False] * n
    size = [0] * n
    ans = 0

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

    def find_centroid(v, p, total):
        for u in graph[v]:
            if u != p and not removed[u]:
                if size[u] > total // 2:
                    return find_centroid(u, v, total)
        return v

    def collect(v, p, d, arr):
        if d > k:
            return
        arr.append(d)
        for u in graph[v]:
            if u != p and not removed[u]:
                collect(u, v, d + 1, arr)

    def decompose(v):
        nonlocal ans

        total = calc_size(v, -1)
        c = find_centroid(v, -1, total)

        freq = [0] * (k + 1)
        freq[0] = 1

        for u in graph[c]:
            if removed[u]:
                continue

            distances = []
            collect(u, c, 1, distances)

            for d in distances:
                if d <= k:
                    ans += freq[k - d]

            for d in distances:
                if d <= k:
                    freq[d] += 1

        removed[c] = True

        for u in graph[c]:
            if not removed[u]:
                decompose(u)

    decompose(0)
    print(ans)

if __name__ == "__main__":
    solve()

The adjacency list stores the tree because every vertex only needs access to its neighbors. The removed array represents vertices that have already become centroids and should not be included in later recursive components.

The first pair of functions, calc_size and find_centroid, locate the centroid of the current component. The size calculation ignores removed vertices because the decomposition is performed independently on each remaining component.

The collect function gathers distances from a centroid into one child subtree. It stops after distance k because longer paths can never contribute to the answer.

The counting phase uses freq as a map from distance to the number of already processed vertices. When processing a new subtree, querying freq[k - d] counts all pairs whose total distance through the centroid is exactly k. Only after querying do we add the subtree distances into freq, preventing pairs from the same subtree from being counted at this level.

Python integers automatically handle the large answer size, which can be close to n². The distance arrays have size k + 1, so there is no hidden memory growth from large paths.

Worked Examples

For the first sample:

A centroid can be vertex 2.

Centroid	Processed distance counts	New subtree distances	Added pairs
2	`{0:1}`	`[1,2]` from vertices 1 and 3 branch	1
2	`{0:1,1:1,2:1}`	`[1]` from vertex 5	2

The total becomes 4 after all recursive components are processed. The trace shows why subtrees are queried before insertion: vertices from different branches form paths through the centroid.

For the second sample:

Choosing the middle vertex as a centroid gives:

Centroid	Processed distance counts	New subtree distances	Added pairs
3	`{0:1}`	`[1,2]`	1
3	`{0:1,1:1,2:1}`	`[1,2]`	1

The answer is 2. This demonstrates that paths on opposite sides of the centroid are counted together while paths inside one side are deferred to recursion.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Each centroid level processes all vertices in its component, and the decomposition depth is logarithmic.
Space	O(n)	The tree, auxiliary arrays, and recursion stack require linear memory.

The constraints allow n = 50000, where quadratic pair enumeration is impossible. Centroid decomposition keeps the total work close to linear per decomposition level and fits comfortably within the required limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    solve()

    result = sys.stdout.getvalue()

    sys.stdin = old_stdin
    sys.stdout = old_stdout
    return result

assert run("""5 2
1 2
2 3
3 4
2 5
""") == "4\n", "sample 1"

assert run("""5 3
1 2
2 3
3 4
4 5
""") == "2\n", "sample 2"

assert run("""1 1
""") == "0\n", "single vertex"

assert run("""3 1
1 2
2 3
""") == "2\n", "distance one"

assert run("""5 2
1 2
2 3
3 4
4 5
""") == "3\n", "chain distance two"

Test input	Expected output	What it validates
Single vertex	0	Handles the smallest tree without edges
Three-node chain with k=1	2	Checks direct edge counting
Five-node chain with k=2	3	Checks path counting without duplicate pairs
Provided samples	4 and 2	Confirms standard cases

Edge Cases

For a tree containing only one vertex, the centroid decomposition starts at that vertex, counts only the centroid itself at distance zero, and finds no pair. The input 1 1 correctly produces 0.

For k = 1, the algorithm counts pairs of adjacent vertices. Since every edge has distance one, every edge is discovered when its endpoints are separated at some centroid level. The input:

3 1
1 2
2 3

is handled by finding two valid pairs and returning 2.

For a path-shaped tree:

the algorithm never counts the same pair twice. A pair such as (1,3) is counted only when a centroid lies on the path between them and separates the two vertices into different processed parts. If they remain together, recursion handles them later.

The implementation also handles cases where k is larger than the tree diameter. All collected distances stop contributing once they exceed k, and no frequency lookup can find a matching distance, so the answer remains zero.