Let's think of the network of cities and highways as a tree rooted at node ~A~. We'll then approach the problem with dynamic programming on this tree. Let ~DP[i][a][b][x]~ be the minimum number of distinct nodes which must be visited by Bo Vine in node ~i~'s subtree such that:

• All movie theatres in node ~i~'s subtree will be visited by somebody
• If ~a = 1~, the Head Monkey will visit node ~i~, and otherwise if ~a = 0~, she will not
• If ~b = 1~, Bo Vine will visit node ~i~, and otherwise if ~b = 0~, he will not
• The Head Monkey will visit ~x~ distinct nodes in node ~i~'s subtree

Starting from some initial node, visiting ~k~ distinct nodes (including the initial one), and then returning to that initial node requires ~2 \times (k-1)~ hours. Therefore, for a given ~b~ value representing whether or not Bo Vine will visit the root (node ~A~), and a given ~x~ value representing how many distinct nodes the Head Monkey will visit in total, ~2 \times (\max(x, DP[A][1][b][x])-1)~ is the total number of hours that will be required. If these ~DP~ values could be computed, then we could consider all ~\mathcal O(N)~ possible pairs of ~(b, x)~ and take the smallest of their corresponding times.

What remains is computing the ~DP~ values, which we'll do recursively starting from the root. The general approach will be a standard one – for each child ~c~ of a given node ~i~, we'll consider all possible triples of ~(a_1, b_1, x_1)~ for node ~i~ and all possible triples of ~(a_2, b_2, x_2)~ for node ~c~, and update node ~i~'s ~DP~ values based on sums of the form ~DP[i][a_1][b_1][x_1]+DP[c][a_2][b_2][x_2]~. The details of interest concern exactly which states and transitions are actually valid, and are as follows:

• If ~C_i = 1~ and ~a = b = 0~, then the state is invalid (node ~i~'s movie theatre must be visited)
• If ~i = B~ and ~b = 0~, then the state is invalid (Bo Vine must visit node ~B~)
• If ~a_1 = 0~ and ~a_2 = 1~, then the transition is invalid (the Head Monkey can't reach node ~c~ without passing downwards through node ~i~)
• If ~B~ is within ~c~'s subtree, ~b_1 = 1~, and ~b_2 = 0~, then the transition is invalid (Bo Vine can't reach node ~i~ without passing upwards through node ~c~)
• If ~B~ is not within ~c~'s subtree, ~b_1 = 0~, and ~b_2 = 1~, then the transition is invalid (Bo Vine can't reach node ~c~ without passing downwards through node ~i~)

To implement the above checks, we can pre-compute whether or not ~B~ is within each node ~i~'s subtree in a total of ~\mathcal O(N)~ time. Given that, the overall time complexity of this dynamic programming approach comes out to ~\mathcal O(N^3)~.