This problem may seem intimidating at first, so let's identify some preliminary clues. First of all, note that the query limit of ~3000~ is quite close to ~300 \log 300~ with moderate overhead. If we recall algorithms which use ~\mathcal{O}(N \log N)~ operations (e.g. merge sort), we may discover the possibility of a divide-and-conquer based approach being applied here.

Let's assume we have 2 disjoint graphs ~G_1~ and ~G_2~ such that both ~G_1~ and ~G_2~ are acyclic. Now, if we introduce some directed edges between ~G_1~ and ~G_2~, how can we ensure that the resulting graph is also acyclic? Well, we can simply direct all the edges such that they lead from ~G_1~ to ~G_2~! Bringing this back to the task at hand, if we query the set of nodes in ~G_2~ as set ~A~ and the set of nodes in ~G_1~ as ~B~, we can just flip the direction of all the edges at the returned indices to ensure that all edges between ~G_1~ and ~G_2~ lead from ~G_1~ to ~G_2~.

Now that we have a method to merge two DAGs into a single larger DAG, we can apply divide-and-conquer on the set of nodes to obtain a working solution using ~\mathcal{O}(N \log N)~ operations.

Alternate Solution

We devise a solution that orients all the edges from a smaller number to a larger number, for which it is easy to show acyclicity using monovariants. Consider the binary representations of the numbers from ~1~ to ~N~. These can each be expressed in ~\mathcal{O}(\log N)~ bits. Furthermore, for each pair of numbers ~i \neq j~ in this range, they must differ in at least one of the ~\mathcal{O}(\log N)~ bits.

Now, for the ~i~-th of the ~\mathcal{O}(\log N)~ bits, let's make a query where a node belongs to set ~A~ if its number has the ~i~-th bit equal to ~0~, or it belongs to set ~B~ otherwise. Then, for each edge that goes from set ~A~ to set ~B~ or vice versa from the input, we flip it if it goes from a larger number to a smaller number (we can implicitly work out the orientations of these edges from the list of indices received from Siesta).

Since each pair ~i \neq j~ must differ by at least one bit, we know that we would have covered all possible pairs (and thus all possible edges) after the ~\mathcal{O}(\log N)~ queries. Since each query includes exactly ~N~ nodes, this solution also uses ~\mathcal{O}(N \log N)~ operations.

Thanks to Nils_Emmenegger for discovering this solution.