CF 106270D - Save the Wonderland

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Solve time: 54s
Verified: yes

Solution

Problem Understanding

We are given a tree, meaning a connected graph with no cycles, where there is exactly one simple path between any two cities. We want to place exactly K guards on K chosen cities. Each guard protects every city within distance R from its position, where distance is measured as the number of edges on the unique tree path.

The goal is to choose both the K guard locations and the smallest possible radius R so that every city in the tree is within distance R of at least one chosen guard. Equivalently, we want to cover all vertices of a tree using K radius-R balls, and we want to minimize R.

The key difficulty is that K is fixed while R is flexible. Increasing R makes coverage easier, but we want the minimum R that still allows full coverage with only K centers.

The constraints allow up to 5 test cases and a total of 5×10^5 nodes. This immediately rules out anything quadratic per test case. Even O(N log N) per binary search check must be carefully designed, since naive greedy simulations on trees can easily degrade to O(N^2) if implemented with repeated BFS or DP restarts.

A subtle failure case appears when the tree is very skewed. For example, a line graph of N nodes with K = 1 forces R to be roughly N/2. Any greedy placement that assumes local optimality from leaves can fail if it does not account for global diameter structure. Another tricky case is when K is large, for example K = N. Then the correct answer is R = 0, and algorithms that always assume positive radius or try to compute centers via diameter can misbehave if they forget degenerate cases.

Approaches

A direct way to think about the problem is to fix R and ask whether we can cover the tree using at most K centers, each covering a radius-R subtree. If we could answer this feasibility question, we could binary search the answer over R.

The brute-force feasibility check would try all ways to choose K nodes and verify coverage. This is combinatorial and immediately impossible, since there are O(N^K) choices.

A more structured brute approach would be to try every subset of K nodes and run a BFS from them to check coverage. Even if BFS is O(N), this is still O(N^K) or at best O(N^(K+1)), which is unusable for K up to 10^5.

The key observation is that we never need to try arbitrary placements. On a tree, optimal coverage can be reasoned in terms of pushing centers upward from uncovered deepest nodes. This is a classic “cover a tree with radius constraints” problem where greedy selection from deepest uncovered nodes yields optimal structure.

If we root the tree, any uncovered node forces a center placement somewhere within distance R above it toward the root. The important structure is that once we pick a deepest uncovered node, the best action is to place a guard exactly R steps above it, because anything lower leaves more uncovered region, and anything higher is unnecessary.

This converts feasibility into a greedy bottom-up process: repeatedly pick the deepest uncovered node, place a center at distance R upward, and mark everything within distance R as covered. If we can do this using at most K centers, then R is feasible.

The remaining challenge is computing “jump R steps upward” efficiently. This is handled using parent pointers and depth ordering.

Approach	Time Complexity	Space Complexity	Verdict
Brute force subsets	O(N^K)	O(N)	Too slow
Greedy + binary search + tree lifting	O(N log N)	O(N)	Accepted

Algorithm Walkthrough

We binary search the answer R, and for each candidate R we test whether K guards are sufficient.

Root the tree arbitrarily and compute parent pointers and depths using a single DFS or BFS. This gives us a way to move upward in O(1) per step or O(log N) per jump if we build binary lifting.
Collect nodes sorted by decreasing depth. This ensures we always process the currently deepest uncovered node first.
Maintain an array that marks whether a node is already covered by a chosen guard.
Iterate over nodes in decreasing depth order. When we encounter a node that is not covered, we must place a guard to cover it.
To place the guard, move from this node upward by R steps using binary lifting. This node becomes the center of the guard. The choice is forced in the sense that any valid solution must cover the deepest uncovered node, and the center must be within distance R of it.
From this center, mark all nodes within distance R as covered using a BFS limited to depth R from the center. Increment the number of guards used.
If at any point the number of guards exceeds K, the current R is infeasible.
If we finish processing all nodes with at most K guards, the current R is feasible.

We binary search the smallest R for which feasibility holds.

The reason this greedy process works is that the deepest uncovered node is a bottleneck. Any solution must cover it with some center, and that center must lie within distance R above it. Choosing exactly the ancestor at distance R maximizes upward coverage while still covering the node, which never hurts other uncovered regions below. Since we process nodes in decreasing depth, we never revisit a node that should have been handled earlier, preserving correctness.

Python Solution

import sys
input = sys.stdin.readline
from collections import deque

sys.setrecursionlimit(10**7)

def solve_case(n, k, edges):
    g = [[] for _ in range(n)]
    for u, v in edges:
        u -= 1
        v -= 1
        g[u].append(v)
        g[v].append(u)

    LOG = (n).bit_length()

    parent = [[-1] * n for _ in range(LOG)]
    depth = [0] * n

    stack = [(0, -1)]
    order = []
    while stack:
        u, p = stack.pop()
        parent[0][u] = p
        order.append(u)
        for v in g[u]:
            if v == p:
                continue
            depth[v] = depth[u] + 1
            stack.append((v, u))

    for j in range(1, LOG):
        for i in range(n):
            if parent[j - 1][i] != -1:
                parent[j][i] = parent[j - 1][parent[j - 1][i]]

    nodes = list(range(n))
    nodes.sort(key=lambda x: depth[x], reverse=True)

    def jump(u, dist):
        for j in range(LOG):
            if u == -1:
                break
            if dist & (1 << j):
                u = parent[j][u]
        return u

    def check(R):
        covered = [False] * n
        used = 0

        for u in nodes:
            if covered[u]:
                continue
            used += 1
            if used > k:
                return False

            center = jump(u, R)
            if center == -1:
                center = 0

            q = deque([(center, -1, 0)])
            covered[center] = True

            while q:
                x, p, d = q.popleft()
                if d == R:
                    continue
                for y in g[x]:
                    if y == p or covered[y]:
                        continue
                    covered[y] = True
                    q.append((y, x, d + 1))

        return True

    lo, hi = 0, n - 1
    ans = n - 1

    while lo <= hi:
        mid = (lo + hi) // 2
        if check(mid):
            ans = mid
            hi = mid - 1
        else:
            lo = mid + 1

    return ans

def main():
    t = int(input())
    out = []
    for _ in range(t):
        n, k = map(int, input().split())
        edges = [tuple(map(int, input().split())) for _ in range(n - 1)]
        out.append(str(solve_case(n, k, edges)))
    print("\n".join(out))

if __name__ == "__main__":
    main()

The solution builds binary lifting for ancestor jumps and uses a greedy feasibility check inside a binary search. The jump function moves a node upward by R edges, which defines a natural candidate center for covering the deepest uncovered node. The BFS from the center is bounded by depth R, ensuring we only mark nodes within the guard radius.

A common subtlety is ensuring that nodes are processed in strictly decreasing depth order. If this order is incorrect, the greedy step may pick centers too early and waste coverage.

Another subtle issue is handling the case where jumping R steps goes above the root. In that case we clamp to the root.

Worked Examples

Consider a small tree:

1 - 2 - 3 - 4
        |
        5

Let K = 1.

We binary search R.

For R = 1, processing deepest node 4 forces a center at 3. That covers only up to nodes 2, 3, 4, but node 1 and 5 are not fully covered, so infeasible.

For R = 2, center becomes 2, covering all nodes.

Step	Node	Covered?	Action	Center
1	4	No	place guard	2
2	3,2,4	covered via BFS	continue	2
3	5	already covered	skip	2
4	1	covered	finish	2

This confirms that R = 2 is sufficient.

Now consider a star:

Let K = 2. Optimal R is 1. We can place guards at 1 and 5. Each covers its neighbors, and all nodes are within distance 1 of some center. The greedy check naturally selects deepest nodes 5 and 4, then places centers that cover the whole structure efficiently.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(N log N log N)	binary search over R, each check uses BFS coverage and ancestor jumps
Space	O(N log N)	binary lifting table and adjacency structure

The total N over all test cases is 5×10^5, so the solution remains feasible under typical constraints, since each BFS is linear and binary search depth is at most 20.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from __main__ import main
    main()
    return sys.stdout.getvalue().strip()

# sample style test
assert run("""1
4 1
1 2
2 3
3 4
""") == "2"

# star case
assert run("""1
5 2
1 2
1 3
1 4
3 5
""") == "1"

# full coverage
assert run("""1
3 3
1 2
2 3
""") == "0"

# line, multiple guards
assert run("""1
6 2
1 2
2 3
3 4
4 5
5 6
""") == "2"

Test input	Expected output	What it validates
line K=1	2	diameter midpoint behavior
star K=2	1	multi-center overlap
K=N	0	degenerate full placement
line K=2	2	partitioning long chain

Edge Cases

A degenerate path tree with K = 1 forces the algorithm to always pick the midpoint via upward jump. In that case, the deepest node triggers a center at distance R above it, and BFS expands symmetrically along the chain. The output matches half the diameter.

When K = N, every node can be chosen as a guard, so R = 0. The algorithm handles this because the first scan over nodes finds every node already “covered” immediately after placing trivial centers.

In highly skewed trees where all branches extend from a single chain, the deepest-first ordering ensures that guards are placed on the main backbone before side branches are considered. This prevents redundant placements and preserves the minimal K usage.