There must always exist an optimal circle which is tangent to at least one trainer ~i~'s line (it can't be better to choose a circle tangent to no lines, as it can always be shifted until it is tangent to one). We'll consider each possible line ~i~ to place the circle tangent to. Given such a line, the circle can then lie on either of the two sides of the line, and we'll consider both possible sides.

Given a line ~i~ and one of its sides, the set of possible circle centers forms a line ~c~ running parallel to line ~i~, ~R~ units away from it. We'll compare this line ~c~ against each trainer ~j~'s line:

• If line ~j~ is parallel to line ~c~, then a circle centered anywhere on line ~c~ will either always or never intersect with line ~j~, depending on whether the distance between those two lines is greater than ~R~. Let ~a~ be the sum of ~P~ values corresponding to all such lines ~j~ which will always be intersected (note that these always include line ~i~ itself).
• Otherwise, if line ~j~ is not parallel to line ~c~, then there exists a single interval of circle centers along line ~c~ for which the circle will intersect with line ~j~. This interval is centered at the intersection of lines ~c~ and ~j~, and its boundaries may be computed with some trigonometry. In particular, we'll want to imagine that line ~c~ is a number line (with an arbitrary origin), and represent each interval's start and end points as scalar points along it.

The number of Pokémon trapped by centering the circle at a given point along line ~c~ is then equal to ~a~ plus the sum of ~P~ values corresponding to all intervals overlapping with that point. The maximum possible value of this expression can be found by performing a line sweep along line ~c~.

For a given line ~c~, comparing it to all of the trainers' lines takes ~\mathcal O(N)~ time, and then performing a line sweep on the resulting intervals takes ~\mathcal O(N \log N)~ time. We'll repeat this whole process twice for each trainer ~i~'s line (once per side) to determine the maximum number of Pokémon which can be trapped by any possibly optimal circle, bringing the total time complexity up to ~\mathcal O(N^2 \log N)~.