I solved this using two binary indexed trees, longest running time being 0.28s. Let [L1,R1], [L2,R2] be intervals, with L1 <= R1, L2 <= R2. To check whether they overlap or not, we simply check if ((R2 < L1) || (L2 > R1)) holds. If this is true, then they are NOT overlapping, otherwise they overlap.

Let the set of all N intervals be given, say [A1,B1] ... [AN,BN], for each player interval [C,D] we want to count the number of NON-overlapping intervals from [Ai,Bi]'s, then the overlap count can be incremented by (N - (non-overlapping count)). We can use a binary indexed tree to preprocess the [Ai,Bi]'s to store the number of Bi's occurred BEFORE a given position, and another BIT tree for the number of Ai's occurring AFTER a given position.

Then, when the player interval [C,D] is given, simply use the BIT trees above to determine the number of Ai's greater then D, and the number of Bi's less then C.