2020/round2/circular_circles_sol.md

We can observe that the weight of the minimum spanning tree is equal to the sum of all edge weights in the graph (Quantity 1), minus the weight of one removed edge per circle (Quantity 2), minus the weight of one additional removed edge somewhere in the graph (Quantity 3).

For Quantity 3, the additional removed edge may either be any inter-circle edge, or potentially a second edge within some circle. There are two categories of circles:

• If (X_i = Y_i), no two of its edges can be validly removed (without disconnecting the graph).
• Otherwise, if (X_i \ne Y_i), the circle can be considered to have two "halves" (the paths connecting its (X_i)th and (Y_i)th nodes clockwise and counterclockwise), such that up to one edge may be validly removed from each half (without disconnecting the graph).

For each circle (i), we'll define three values of interest:

• Let (R_i) be the maximum weight of any edge within the circle. This will help with computing Quantity 2.
• Let (P_i) be the maximum possible combined weight of any validly-removable pair of edges within the circle (if any). If (X_i = Y_i) (meaning that only one of its edges may be removed), we'll let (P_i = R_i) instead.
• Let (D_i = P_i - R_i). The represents the additional amount by which the minimum spanning tree's weight can be decreased if two edges are removed from the circle, and will help with computing Quantity 3.

Then, for each circle (i), we'll maintain:

• An ordered multiset of all edge weights within it (used to compute (R_i)).
• If (X_i \ne Y_i), two additional ordered multisets containing all edge weights in each of its halves (used to compute (D_i)).

On top of that, we'll globally maintain:

• The sum of all edge weights in the graph (Quantity 1).
• The sum of all circles' (R_i) values (equivalent to Quantity 2).
• An ordered multiset of all circles' (D_i) values, and another containing all inter-circle edge weights (combined to compute Quantity 3).

Each event (and initial edge) requires us to update just a constant number of the above values and multisets, each taking logarithmic time (provided that the multisets are implemented using BBSTs). Therefore, this solution has a time complexity of (O((NM + E) (log(N) + log(M)))).