Subtask 1

We can model each student as a 3-dimensional coordinate ~(a_i, b_i, c_i)~. We will first "plane sweep" (like line sweep, but in 3 dimensions) through the x dimension in non-increasing order. We sweep with a plane parallel to the yz-plane, having its x-coordinate denote the current ~A~ being considered. This means that any coordinate with ~a_i \le A~ will get covered. We can treat the remaining points with ~a_i > A~ as 2-dimensional coordinates ~(b_i, c_i)~. Since we are sweeping in non-increasing order, our algorithm will add new ~(b_i, c_i)~ points into consideration and never remove any old points. Let ~f(y)~ be the minimum value of ~C~ we need to cover all the points if we set ~B=y~ at the current moment. When we add a point ~(b_i, c_i)~ into consideration, all ~f(y)~ with ~y < b_i~ will get updated to ~\max(f(y), c_i)~. For the first subtask, it is sufficient to loop to update all ~y < b_i~. We can then loop through all ~y~ to find the minimum ~A+y+f(y)~ (we set ~B=y~ and ~C=f(y)~). Our answer is the minimum of all ~A+y+f(y)~. This costs ~\mathcal{O}(N)~ per point for a total time complexity of ~\mathcal{O}(N^2)~.

Subtask 2

Notice that the function ~f(y)~ is monotonically non-increasing. Instead of looping to update and query, we can use a data structure such as a segment tree or a balanced binary search tree (BBST). ~\mathcal{O}(N \log N)~ segment tree solutions should pass with ease. Some optimization may be required to get BBST solutions or ~\mathcal{O}(N \log^2 N)~ solutions to pass.