Let \(F(t)\) equal \(true\) if it's possible to light the fuse within at most \(t\) seconds, and equal to \(false\) otherwise. If \(F(t_1) = true\), this implies that \(F(t_2) = true\) for every \(t_2 \ge t_1\). This means that it's possible to binary search for the smallest value of \(t\) such that \(F(t) = true\), which will be our answer. This value must be positive and no greater than \(D_P + D_Q\). What remains is efficient evaluation of \(F(t)\).

Let \(E_i\) be the set of log cluster positions which the \(i\)th log driver will set explosive charges at. If \(E_i\) is non-empty, we can observe that the number of seconds the \(i\)th log driver requires is equal to \(\text{min}(|\text{min}(E_i) - P_i|, |\text{max}(E_i) - P_i|)\) \(+\) \((\text{max}(E_i) - \text{min}(E_i))\), where \(\text{min}(E_i)\) and \(\text{max}(E_i)\) are the smallest and largest positions in \(E_i\), respectively.

This observation may be turned into a key insight — if we consider the log drivers and the log clusters each sorted in increasing order of position, then it's optimal for \(E_{1..N}\) to correspond to a partioning of the log clusters into a sequence of contiguous subarrays handled by one log driver each, in the same relative order. Interleaving two log drivers' \(E\) sets (such that either of them no longer consists of a subarray of log clusters) can only increase the time required, as can swapping two \(E\) sets (such that an earlier log driver sets explosive charges at a later subarray of log clusters).

Based on this insight, we can iterate over the log drivers in increasing order of position, while maintaining a non-decreasing pointer into the sorted list of log clusters and greedily allocating as many of the next log clusters as possible to the current log driver (without them requiring more than \(t\) seconds to handle those log clusters). By the time all \(N\) log drivers have been considered, we must have \(F(t) = true\) if and only if all \(M\) log clusters have been allocated to them.

Evaluating \(F(t)\) takes just \(O(N+M)\) time, assuming that the \(P\) and \(Q\) values have already been sorted. As such, the overall time complexity of this solution comes out to \(O(N log(N) + M log(M) + (N+M) log(D_P + D_Q))\).

Though we initially believed that the above solution extends to the case in which \(S > 0\), it does not, as it is then no longer necessarily the case that the partitions of logs clusters should be assigned to log drivers in the same relative order. For example, consider the counterexample (pointed out by competitor Scott Wu) in which there are 2 log drivers at positions \(20\) and \(25\), 3 log clusters at positions \(5\), \(10\), and \(11\), and \(S = 10\). In this case, the first log cluster should be handled by the second log driver to achieve the optimal answer of 30.

Because this error unfortunately only came to light during the round, we altered the constraints at that point such that \(S\) is always equal to \(0\) rather than being positive. Since everybody who received an Accepted verdict before that point had written a solution trivially applicable to \(S = 0\) and consistent with our intended solution, we kept those Accepted verdicts, while rejudging all submissions which had previously received Wrong Answer verdicts.