First, we note that there are up to ~2N~ significant points that Fax can visit: all ~N~ checkpoints, and all points that are ~S~ meters below a checkpoint. If ~S=0~ or the point has a negative altitude, we ignore the second point. For each point, we store the number of checkpoints Fax gains if he passes through this point. As such, each point will either have a value of ~1~ (checkpoint with nothing above, empty point with checkpoint above) or ~2~ (checkpoint with checkpoint above). For the ~i^{th}~ point, let this value be ~v_i~.

Next, we can use dynamic programming to compute the answer. For each point, we want to know the maximum number of checkpoints Fax could have passed through after going through the current point and performing a loop. For the ~i^{th}~ point, let this value be ~a_i~. We can see that ~a_i = \max(a_j + v_i)~ for all points ~j~ with a distance less than point ~i~ that can be reached by ascending or descending.

The answer is the maximal ~a_i~.

Time Complexity: ~\mathcal{O}(N^2)~