IMO Shortlist, 2019

We will show Alice needs to ask at most $4 n-7$ questions. Her strategy has the following phases. In what follows, $S$ is the set of towns that Alice, so far, does not know to have more than one outgoing road (so initially $|S|=n$ ).

Phase 1. Alice chooses any two towns, say $A$ and $B$. Without loss of generality, suppose that the King of Hearts' answer is that the road goes from $A$ to $B$.
At the end of this phase, Alice has asked 1 question.

Phase 2. During this phase there is a single (variable) town $T$ that is known to have at least one incoming road but not yet known to have any outgoing roads. Initially, $T$ is $B$. Alice does the following $n-2$ times: she picks a town $X$ she has not asked about before, and asks the direction of the road between $T$ and $X$. If it is from $X$ to $T, T$ is unchanged; if it is from $T$ to $X, X$ becomes the new choice of town $T$, as the previous $T$ is now known to have an outgoing road.
At the end of this phase, Alice has asked a total of $n-1$ questions. The final town $T$ is not yet known to have any outgoing roads, while every other town has exactly one outgoing road known. The undirected graph of roads whose directions are known is a tree.

Phase 3. During this phase, Alice asks about the directions of all roads between $T$ and another town she has not previously asked about, stopping if she finds two outgoing roads from $T$. This phase involves at most $n-2$ questions. If she does not find two outgoing roads from $T$, she has answered her original question with at most $2 n-3 \leqslant 4 n-7$ questions, so in what follows we suppose that she does find two outgoing roads, asking a total of $k$ questions in this phase, where $2 \leqslant k \leqslant n-2$ (and thus $n \geqslant 4$ for what follows).
For every question where the road goes towards $T$, the town at the other end is removed from $S$ (as it already had one outgoing road known), while the last question resulted in $T$ being removed from $S$. So at the end of this phase, $|S|=n-k+1$, while a total of $n+k-1$ questions have been asked. Furthermore, the undirected graph of roads within $S$ whose directions are known contains no cycles (as $T$ is no longer a member of $S$, all questions asked in this phase involved $T$ and the graph was a tree before this phase started). Every town in $S$ has exactly one outgoing road known (not necessarily to another town in $S$ ).

Phase 4. During this phase, Alice repeatedly picks any pair of towns in $S$ for which she does not know the direction of the road between them. Because every town in $S$ has exactly one outgoing road known, this always results in the removal of one of those two towns from $S$. Because there are no cycles in the graph of roads of known direction within $S$, this can continue until there are at most 2 towns left in $S$.
If it ends with $t$ towns left, $n-k+1-t$ questions were asked in this phase, so a total of $2 n-t$ questions have been asked.

Phase 5. During this phase, Alice asks about all the roads from the remaining towns in $S$ that she has not previously asked about. She has definitely already asked about any road between those towns (if $t=2$ ). She must also have asked in one of the first two phases about
at least one other road involving one of those towns (as those phases resulted in a tree with $n>2$ vertices). So she asks at most $t(n-t)-1$ questions in this phase.
At the end of this phase, Alice knows whether any town has at most one outgoing road. If $t=1$, at most $3 n-3 \leqslant 4 n-7$ questions were needed in total, while if $t=2$, at most $4 n-7$ questions were needed in total.