"board" has an solution: first, you need to convert the string to path consisting of "1"s and "2"s only; for that, use stack, in which you remember sequences of consecutive ones and twos; "U" is removing the last number from the last sequence, "1" and "2" is adding it to the last sequence, "L" is removing all "1"s from the end (let there be K of them), replacing the last "2" by "1" and adding K twos, and similarly for "R". It's clear that the best solution is going greedily upwards from both nodes to some (the same) level, and then just moving on that level; also, it always fits into an integer, because the level we go to is at most the root (length equal to sum of depths of initial nodes). So, try all levels from the root downwards, re-calculate the horizontal distance between that level's ancestors of the initial nodes (it's just a number in binary), if it's clearly too large then break (to avoid overflow), and choose the best answer of "horizontal distance + sum of levels of initial nodes — 2*current level".

Adriatic: draw a picture of what areas on the grid contain only islands of distance d. For d = 1, it's 2 non-intersecting rectangles touching the north-west and south-east corners of the map. For d = 2, it's the whole map minus 2 rectangles touching the other 2 corners, and not intersecting, and it's easy to show (by drawing it) that for any larger d, it's just some 2 smaller rectangles. How those 2 rectangles change depends only on the island with the largest/smallest x/y coordinates of those with distances d and less, which we can pre-compute in the same way as prefix sums (60pts is simulating this progress, then) and in a similar way — dynamic programming, in fact — we can count the sum of distances for any rectangle. Time is O(2500² + N).

Codes: board (be warned, it contains other stuff like bruteforce and random number generator, because no feedback), adriatic

For Slovakia the answer is no. Well, yes, Eduard is currently the best one we have, but the rest of the IOI 2013 team did not take part. We usually send younger students to the CEOIs. All four Slovak contestants at CEOI 2013 are eligible to compete at IOI 2014 (if they succeed to qualify, of course).