Once again, we can see the system as a graph with ~N~ nodes (cisterns). Due to the fact that there are ~N-1~ edges and no cycles, we can also see that the graph is a tree. You can read more about properties of trees from this article. In this case, we can see that cistern ~1~ is the root of the tree, and that cistern ~C_i~ is the parent of node ~i~. In addition, the problem conveniently guarantees that node ~i~ is the parent of node ~j~ if and only if ~i < j~. Everything about it points towards a solution using the classic dynamic programming (DP) on a tree approach. Many articles for DP on a tree can be found with a quick search online.

Let our dynamic programming state ~DP[i][j]~ represent the maximum amount of chocolate milk which can flow into cistern ~i~ (from all sources, direct or indirect), after ~j~ of the pipes in ~i~'s subtree (not including the pipe leading down from it, if any) have been upgraded. The final answer (the maximum amount of chocolate milk which can flow into cistern ~1~ after ~K~ pipes have been upgraded) will then be ~DP[1][K]~.

For each cistern ~i~ from ~N~ down to ~1~, we'll compute ~DP[i][0 \dots K]~ all at once based on the DP values of its children. For each of its child ~c~, we can choose to distribute some number ~u~ of pipe upgrades to its subtree, and we can choose to either upgrade to pipe from ~u~ to ~i~ (in which case the amount of flow into ~i~ will be ~DP[c][u]~), or not (in which case it'll be ~\max\{DP[c][u], F_u\}~). If ~i \ge 2~, an additional flow of ~P_i~ will also always be added. We can explore all possible decisions regarding how to distribute pipe upgrades amongst ~i~'s children using secondary DP, letting ~DP2[j][k]~ be the maximum flow from the first ~j~ children with ~k~ pipe upgrades. Letting ~M_i~ be the number of children that cistern ~i~ has, we can then copy over the results into the primary DP, with ~DP[i][k] = DP2[M_i][k]+P_i~ (for ~0 \le k \le K~).

The secondary DP takes ~\mathcal O(M_i \times K^2)~ time to compute for each cistern ~i~. Note that the sum of all ~M_i~ for all ~i = 1 \dots N~ is equal to ~N-1~, since every cistern (except the root) is the child of one other cistern, so this DP (and the whole solution) will take a total of ~\mathcal O(N \times K^2)~ time.