I. Steppe on It

Key Observations

• Fix a candidate response time $R$. Then the question becomes: can we place at most $f$ engines so that every village is within distance $R$ of some engine?

• This feasibility predicate is monotone. If radius $R$ works, then every larger radius also works. Therefore we can binary search the answer.

• For the feasibility check, root the tree arbitrarily. When processing a subtree, only one of two summaries matters:

  • the subtree is already fully covered, and we only need the distance from the root of the subtree to the closest engine inside it;

  • the subtree is not fully covered yet, and we only need the distance to its farthest still-uncovered village.

  • A greedy postorder DFS is enough: whenever the farthest uncovered village can no longer be covered from the parent, we must place an engine at the current node.

  Algorithm

  1. Binary search the minimum feasible radius $R$.

  2. For a fixed $R$, run one DFS from the root.

  3. For a node $u$, after processing all children, maintain:

     • closest_center: the minimum distance from $u$ to an engine already placed in a child subtree that is fully covered;

     • farthest_uncovered: the maximum distance from $u$ to a village in a child subtree that is still uncovered.

     • If \[ \texttt{closest\_center} + \texttt{farthest\_uncovered} \le R, \] then the nearest engine inside the subtree covers every currently uncovered village as well, so the whole subtree becomes fully covered and we return closest_center.

     • Otherwise, if $u$ is not the root and \[ \texttt{farthest\_uncovered} > R, \] then even an engine at the parent would be too far away, so we must place an engine at $u$. We count one engine and return that the subtree is fully covered with nearest-engine distance $0$.

     • If neither of the two situations above applies, we return that the subtree is still uncovered, with value farthest_uncovered.

     • After the DFS, if the root is still uncovered, place one final engine at the root. The check succeeds iff the total number of engines used is at most $f$. enumerate

       Correctness Proof

       We prove that the algorithm returns the correct answer.

       Lemma 1.

       During the feasibility check for radius $R$, if a node $u$ returns the state ``uncovered with value $d$'', then every uncovered village in the subtree of $u$ is at distance at most $d$ from $u$, and at least one such village is at distance exactly $d$.

       Proof.

       This follows directly from the DFS definition. Whenever a child returns uncovered, we propagate the maximum uncovered distance upward by adding the edge length. Thus the stored value is exactly the farthest uncovered distance from the current node. □

       Lemma 2.

       If at a non-root node $u$ we have farthest_uncovered $> R$, then every feasible solution for radius $R$ must place an engine at $u$ or inside the subtree of $u$.

       Proof.

       By Lemma 1 there is an uncovered village in the subtree of $u$ at distance more than $R$ from $u$. Any engine outside the subtree of $u$ must be even farther away, because every path from that village to the outside passes through $u$. Hence no engine above $u$ can cover that village. Therefore some engine must be placed at $u$ or below it. The greedy choice of placing it exactly at $u$ is optimal because it is the highest possible placement that still covers the problematic part of the subtree. □

       Lemma 3.

       If closest_center $+$ farthest_uncovered $\le R$ at node $u$, then all uncovered villages in the subtree of $u$ are in fact already covered by the nearest engine in that subtree.

       Proof.

       Every uncovered village is at distance at most farthest_uncovered from $u$, while the nearest engine is at distance closest_center from $u$. The unique path between them goes through $u$, so their mutual distance is at most the sum of the two values, which is at most $R$. Hence all those villages are actually covered, and the whole subtree is fully covered. □

       Theorem.

       The algorithm outputs the correct answer for every valid input.

       Proof.

       For a fixed radius $R$, Lemma 2 shows that whenever the DFS places an engine, every feasible solution must also place an engine somewhere in that subtree, so the greedy action is necessary. Lemma 3 shows that whenever the DFS declares a subtree fully covered, this is correct. Therefore the DFS computes the minimum number of engines needed for radius $R$.

       Since the feasibility predicate is monotone in $R$, binary search returns the smallest radius for which the DFS uses at most $f$ engines. This is exactly the minimum guaranteed response time. □

       Complexity Analysis

       Each feasibility check is one DFS, so it costs $O(n)$. The answer is found by binary search over an integer range whose upper bound is the total tree length, so the total running time is $O(n \log W)$, where $W$ is that upper bound. The memory usage is $O(n)$.

       Implementation Notes

       • Edge lengths and path lengths require long long.

       • The DFS is recursive in the write-up, but any equivalent postorder implementation is fine. The code roots the tree at village $1$.

       • If the root is still uncovered after processing all children, we must count one additional engine at the root.