Thank you for perfect explanation! Actually, this is the author's solution =)

Interesting, does anybody solved it in another way?

Author's solution is offline: sort queries by time -> solve problem -> sort queries by id -> output. Main idea is to move T_i instead of S_i. You need to store list of validity time intervals S_i..S_i + 1 (S_n..∞ for the last one) for typed numbers (only if 1 ≤ B_i ≤ N). For each interval you can use lower_bound/upper_bound (with delta depending on T_i of current question) to find first query to increase and decrease counter of correct answers and put such events to priority queue (or just sorted array). The last step is to process all requests one by one using +1/-1 events from priority queue.

Online solution exists too, but we decided to make simple and fun contest for our on-site participants :)

We've approached it like this. Let's construct convex hull. After that you calculate answers for all segments of convex hull with two pointers.

Given one segment you want to find the farthest segment (in terms of index in convex hull array) such that you can remove all points between these two segments. You want the rays constructed from the segments (you connect i-th and (i + 1)-th points for one ray and (j + 1)-th and j-th for another, I mean different directions) to intersect. This can be interpreted as "the angle between the rays is strictly less than π", this can be checked with cross product of their vectors.

Several points on the same line don't make difference because you will always have the better answer when you take the earliest clockwise of them. We only solved the case with the whole convex hull being on one line with separate if.

Hope the image tells it clearer. :D