Interactive Graph (Simple Version)

Research basis: I checked the official statement and the Codeforces Round 1079 editorial. The official C1 solution uses the binary-search-over-path-index approach described here; the reasoning below is original.

First, do not get baited by the hack format. In the actual interactive solution, your program reads only $t$ and then each $n$ from the interactor. The hidden m and edge list are for the interactor/hacker input, not for your solution to read. Reading m is a one-way ticket to WA city.

For any vertex $u$, all paths whose first vertex is $u$ form one contiguous block in the global lexicographic ordering. Blocks are ordered by the starting vertex: all paths starting at $1$, then all paths starting at $2$, and so on.

Inside the block for $u$, the path $[u]$ comes first, because it is a proper prefix of every longer path starting at $u$. After that, paths are grouped by their second vertex. So if $u$ has outgoing neighbors $v_1 < v_2 < ... < v_d$, the block for $u$ looks like:

$[u]$, then all paths starting with $[u,v_1]$, then all paths starting with $[u,v_2]$, and so on.

The first path in the group for neighbor $v$ is exactly $[u,v]$, since that direct two-vertex path is a prefix of every longer path $[u,v,...]$. Therefore, if we can find the first path whose source is $u$ and whose second vertex is larger than the last neighbor we found, we get the next outgoing edge of $u$.

Fix a source $u$ and let $prev$ be the largest outgoing neighbor of $u$ already discovered. Initially $prev=-1$. Define a queried path $P$ to be after the skipped prefix if one of these holds:

• $P$ does not exist, meaning the queried index is past the end of the path list.
• $P_1 > u$.
• $P_1 = u$, $|P| > 1$, and $P_2 > prev$.

Everything else is still before or inside the part already consumed: it starts below $u$, it is the singleton $[u]$, or it starts with $[u,v]$ where $v <= prev$. Because of the ordering above, this predicate is monotone over $k$: false for a prefix of indices, then true forever. No weird interleaving happens; lexicographic order does the bookkeeping for us.

So we binary search $k$ from $1$ to $2^{30}$ for the first index where the predicate becomes true. If the returned path starts at $u$ and has length at least $2$, then its first two vertices are $u$ and the smallest not-yet-found neighbor $v>prev$, so edge $(u,v)$ exists. Add it, set $prev=v$, and repeat.

If the returned path is empty or no longer starts at $u$, there is no remaining outgoing neighbor from $u$. The first real path of any later starting vertex is its singleton path, so a later block cannot make us add a fake edge for $u$. Still, checking path[0] == u in code is cheap paranoia, and cheap paranoia is usually good CP hygiene.

Correctness proof:

For a fixed vertex $u$, lexicographic order makes all paths starting at $u$ contiguous. Within that block, after the singleton $[u]`, paths are ordered by their second vertex. Therefore, after all groups with second vertex at most$prev$are skipped, the first path with source$u$and second vertex greater than$prev$exists if and only if$u$has an outgoing neighbor greater than$prev$. When it exists, the first such path is$[u,v]$for the smallest such neighbor$v$, so adding$(u,v)$ is correct.

The binary search is valid because the predicate describes a suffix of the sorted path list. Thus it finds exactly the first candidate after the skipped prefix. If that candidate begins with $u$ and has length at least $2$, the previous paragraph proves it gives the next real outgoing edge. If it does not, then no remaining path of the form $[u,v,...]$ with $v>prev$ exists, so all outgoing edges of $u$ have already been found.

By induction over repeated searches for one source $u$, the algorithm finds all outgoing edges of $u$ in increasing order and no fake edges. Running this for every source vertex outputs every edge exactly once.

Each actual edge causes one successful binary search. Each vertex causes one additional failed binary search proving its outgoing list is done. That is $m+n$ searches. A search over $[1,2^{30}]$ uses at most $31$ queries, and we ask one more query at the found index to read the path, so the total is at most $32(m+n)$ queries.

Complexity: at most $32(n+m)$ queries. Local work is $O(30(n+m))$ plus reading returned paths, and memory is $O(n+m)$.