Indeed

It's an unfortunate coincidence, but we planned the nationals before knowing about RMI and at this point we can't do much to avoid it. We always hold the mirror in parallel to the onsite contest, but the tasks will be made available on our training platform soon after the end.

Can you share website or other information about RMI?

I think jamesbamber will win, he is too strong

The ranking is now available at ranking.olinfo.it/2023-pre-oii-mirror/, and the second contest started ~15min ago

it also states that " Every user is allowed to compete (i.e. submit solutions) for a uninterrupted time frame of 3 hours. "

When it is written in the blog that : " When the contest starts, you will see a red button. Click it when you want to start your 5 hour time window! "

It won't be possible as a virtual contest, but we will upload the tasks on training.olinfo.it

This is a known problem, the ranking doesn't pick up new users until we manually do it. I will update it when I get back home.

I couldn't see myself before either, but it updated recently so check now :)

A solution PDF (Italian only, sorry!) has been published at wiki.olinfo.it/it/2023/nazionali

First, in fact, there are $$$O(N \log N)$$$ subgrids (= rectangular areas) which is valid. For example, if the drawing is a "complete grid", the only valid subgrid are $$$2^k \times 2^k$$$ grid, which is only $$$\sum (L-2^k+1)^2 = O(L^2 \log L)$$$, which is $$$O(N \log N)$$$. One can prove that this property holds for any drawing.

The strategy is to enumerate all of them by merging, like the following:

1. The grid is separated into some region. First, compute all of the regions and check if it is all rectangular (if some are not, it's an invalid grid). Let $$$S$$$ be the set of rectangles.
2. Repeat the following while we can make progress:
   • Choose 4 rectangles $$$R_1, R_2, R_3, R_4 \in S$$$. If we can make a larger rectangle by merging $$$4$$$ rectangles which the "center of the cross" is the same, add the larger rectangle to $$$S$$$.
3. Finally, if the entire grid $$$[0, L] \times [0, L] \in S$$$, the grid is valid. Otherwise, the grid is invalid. The reconstruction of the answer when the grid is valid can be done by following:
   • For any rectangle $$$R \in S$$$, record the index of rectangles $$$R_1, R_2, R_3, R_4$$$ that made up $$$R$$$ in Step 2.
   • Then, we can construct the answer by running DFS.

The main part is Step 2, as the naive solution takes $$$O((N \log N)^4)$$$, which is impossible to get 100 points. However, there is a following convenient property on set $$$S$$$:

• Suppose $$$[x_l, x_r] \times [y_l, y_r] \in R$$$ and $$$[x'_l, x'_r] \times [y'_l, y'_r] \in R$$$. If $$$(x_l, x_r, y_l) = (x'_l, x'_r, y'_l)$$$, then $$$y_r = y'_r$$$.
• This property, of course, also holds for the other $$$3$$$ directions.

So, when a new rectangle $$$R$$$ is added to set $$$S$$$, and if the "center of cross" is determined (one of $$$4$$$ choices), the other rectangles $$$R_2, R_3, R_4$$$ for merging, when $$$R_1 = R$$$, can be uniquely determined. This process can be done in $$$O(1)$$$ time if we use hashmap.

Therefore, this problem can be solved in $$$O(N \log N)$$$. The constant factor can be somewhat large, but I think it at least gets 87 points.

They are already out at ranking.olinfo.it/2023-oii