I Yearned For The Mines

The clean way to think about this problem is: stop trying to predict Herobrine. That way lies pain, sadness, and probably a wrong greedy. Instead, reshape the tree into something stupid enough that a deterministic sweep works.

Sweeping a path

Suppose a connected component is a path

$v_1 - v_2 - \cdots - v_k.$

If we check $v_1, v_2, \ldots, v_k$ in this order, Herobrine is guaranteed to be caught if he is in this component.

Why? After checking $v_i$, he cannot be at $v_i$: either he was there and got caught, or he was elsewhere and is not allowed to move into the just-checked vertex. Also, he cannot cross from the unsearched suffix into the searched prefix without stepping through $v_i$. So after the first $i$ checks, vertices $v_1, \ldots, v_i$ are impossible. After $k$ checks, there is nowhere left.

So if the whole graph were a path, $n$ operations of type 1 would solve it.

What we need to build

We will choose some vertices called black vertices and perform operation 2 on all of them first. This deletes every edge incident to a black vertex. After that, every black vertex is isolated, and the non-black vertices form some forest.

If every non-black connected component is a path, then we are done:

1. Destroy every black vertex.
2. Check every black vertex once. They are isolated now.
3. For each non-black path component, check its vertices from one endpoint to the other.

The number of operations is

$n + B,$

where $B$ is the number of black vertices: every vertex is checked once, and every black vertex gets one extra destroy operation. Therefore we only need $B \le \lfloor n/4 \rfloor$.

Now the whole problem becomes: choose at most $\lfloor n/4 \rfloor$ vertices whose removal of incident edges leaves only paths.

The coloring DP

Root the tree at vertex $1$. We process vertices bottom-up and assign one of three colors:

• green: the component inside this subtree that contains this node is a path, and this node is an endpoint that can still connect to its parent.
• yellow: the component inside this subtree that contains this node is already a path, but this node is an internal vertex with two child directions, so it must not connect upward.
• black: we will destroy this vertex, so its parent should ignore this whole connection.

For a node $v$, only its non-black children matter.

Let green_count be the number of green children, and let has_yellow say whether it has a yellow child.

The rules are:

• If has_yellow is true, color $v$ black.
• Else if green_count >= 3, color $v$ black.
• Else if green_count == 2, color $v$ yellow.
• Else color $v$ green.

Why these rules?

A yellow child is already internal in a path. If $v$ stayed non-black, the edge from $v$ to that yellow child would attach a third direction to that child, breaking the path structure. So $v$ has to be black.

If $v$ has at least three green children, then keeping $v$ would connect at least three path branches through $v$. Degree $3$ in a supposed path. Nope. Black it.

If $v$ has exactly two green children, it can connect those two child paths into one longer path, with $v$ as the middle. That makes $v$ yellow.

If $v$ has zero or one green child, then $v$ is still an endpoint of a path that can continue upward, so it is green.

This guarantees that after all black vertices are destroyed, every remaining component has maximum degree at most $2$. Since the original graph is a tree, those components cannot contain cycles, so they are paths. Good. No weird graph theory jump scare here.

Why there are few black vertices

We need the important bound: $B \le \lfloor n/4 \rfloor$.

For every black vertex, we can assign it three private non-black vertices nearby:

• If it became black because it had a yellow child, take that yellow child and the two green children that made that child yellow.
• Otherwise, it became black because it had at least three green children, so take any three of those green children.

These assigned vertices are private. A vertex can only be used by the nearest black vertex above it in this bottom-up structure; once a black vertex appears, ancestors see that child as black and do not reach through it for their own assignment.

So each black vertex accounts for a group of at least four distinct vertices: itself plus three private helpers. Therefore

$4B \le n,$

so

$B \le \lfloor n/4 \rfloor.$

That gives the operation bound:

$n + B \le n + \lfloor n/4 \rfloor = \lfloor 5n/4 \rfloor.$

Constructing the output

After coloring:

1. Output 2 x for every black vertex $x$.
2. Output 1 x for every black vertex $x$.
3. In the graph induced by non-black vertices, every component is a path. For each component, find an endpoint, walk along the path, and output 1 x for each vertex in that order.

The checks over different components can be in any order because the destroy operations disconnected them. If Herobrine is in some component we have not started sweeping yet, he can wiggle around inside it all he wants. Once we start sweeping that path from an endpoint, the path argument traps him anyway.

Correctness proof

First, the path-sweep lemma says that checking a path from one endpoint to the other catches Herobrine if he is in that path. The invariant is that after checking the first $i$ vertices, none of them can contain him, and he cannot cross back into them.

Second, the coloring rules ensure every non-black component is a path. Green nodes have at most one non-black child, yellow nodes have exactly two green child directions and cannot connect upward because their parent becomes black, and black nodes have all incident edges destroyed. Thus no remaining vertex has degree more than $2$.

Third, after all destroy operations, Herobrine is in exactly one component of the remaining forest. Black components are single isolated vertices and are checked directly. Non-black components are paths and are swept from an endpoint, so the path-sweep lemma catches him.

Finally, the charging argument proves there are at most $\lfloor n/4 \rfloor$ black vertices. Therefore the algorithm uses at most $n + \lfloor n/4 \rfloor = \lfloor 5n/4 \rfloor$ operations.

The algorithm is linear per test case: $O(n)$ time and $O(n)$ memory.