The network of locations and roads forms a tree, rooted at its ~1^\text{st}~ node. Anytime Judy or Nick travel down an edge from a node to one of its children, they'll need to travel back up it later, so for convenience we can only count the downwards edge traversals and then multiply the answer by ~2~ at the end.

This problem can be approached with dynamic programming, split into two portions. First, let ~DP1[i]~ be the minimum amount of time required for a single cop to visit all of the required nodes in node ~i~'s subtree (starting and ending at ~i~), with the other cop never entering said subtree. For convenience, let ~DP1[i] = -1~ if ~i~'s subtree contains no nodes with clues. We can initialize ~DP1[i]~ to ~0~, and for each child ~c~ of ~i~ such that ~DP1[c] > -1~, we can add ~DP1[c]+1~ onto ~DP1[i]~, as the cop will need to traverse the edge down from ~i~ to ~c~ and then proceed to visit all of ~c~'s subtree's required nodes. Finally, we should remember to set ~DP1[i]~ back to ~-1~ if ~C_i = 0~ and ~DP1[c]~ was ~-1~ for each child ~c~.

Next, let ~DP2[i][j]~ be the minimum amount of time which must be spent by the first cop within ~i~'s subtree (starting and ending at ~i~), given that the second cop is also contributing by spending ~j~ minutes within ~i~'s subtree (also starting and ending at ~i~), and such that all required nodes in node ~i~'s subtree end up getting visited by at least one of the cops. Note that the answer we're looking for is the minimum value of ~\max(j, DP2[1][j])~ across all possible values of ~j~ between ~0~ and ~N~.

The trickiest part of the algorithm is actually computing the values of ~DP2[i][0 \dots N]~. We can start by initializing ~DP2[i][0 \dots N] = 0~, and then consider each of ~i~'s children ~c~ in turn such that ~DP1[c] > -1~. We'll want to compute a temporary updated copy of the DP array ~DP2'[i][0 \dots N]~ (whose values should first be initialized to infinity), and then copy ~DP2'[i][0 \dots N]~ back into ~DP2[i][0 \dots N]~. Now, there are ~3~ types of options for this child - either the first cop should go down and handle its subtree alone, or the second cop should do so, or both cops should visit it. The first option corresponds to a recurrence of the form ~DP2'[i][j] = \min(DP2'[i][j], DP2[i][j]+DP1[c]+1)~, for all ~\mathcal O(N)~ values of ~j~. The second option indicates ~DP2'[i][j+DP1[c]+1] = \min(DP2'[i][j+DP1[c]+1], DP2[i][j])~. The third option indicates ~DP2'[i][j+k+1] = \min(DP2'[i][j+k+1], DP2[i][j]+DP2[c][k]+1)~ (note that, for each ~j~, we need to also consider all ~\mathcal O(N)~ possible values of ~k~, corresponding to how the cops may split their time in ~c~'s subtree).

The values of ~DP1[i]~ and ~DP2[i][0 \dots N]~ all depend only on the DP values of ~i~'s children, meaning that all of the DP values may be computed recursively starting from node ~1~. For each node ~i~ visited during the recursion, updating the value of ~DP1[i]~ based on each of ~i~'s children takes ~\mathcal O(1)~ time, and updating the values of ~DP2[i][0 \dots N]~ based on each of ~i~'s children takes ~\mathcal O(N^2)~ time. Each node is the child of at most one other node, meaning that the overall time complexity of this algorithm is ~\mathcal O(N^3)~.