Pepero Exchange

For any height $t$, connect only fences of height at most $t$. A pair of cost at most $t$ must stay within one component, and a component of size $s$ contains at most $\lfloor s/2\rfloor$ disjoint pairs. Process fences in increasing height order, pairing as many people inside each component as possible and retaining one unmatched person exactly when its size is odd. Merging two odd components across a fence of height $h$ creates one new pair of cost $h$; other merges create none. The maximum fence height between the two unmatched people is exactly $h$. This construction simultaneously attains the maximum possible number of pairs of cost at most $t$ for every threshold $t$. Consequently its sorted pair costs are each no larger than the corresponding sorted costs of any perfect matching, so their sum is minimum. Sorting and DSU take $O(N\log N)$ time and $O(N)$ space. For subtask 1, subset DP pairs the first remaining person with every possible partner, with an upper bound of $O((2N)^2 2^{2N})$ time.

Solution written by GPT6.1