Hello Codeforces!

Today was Codeforces Round #720 (Div.2), in which I solved problem D — Nastia Plays with a Tree. As I find my solution different from the solutions I heard of after the contest, I will explain it in detail in this blog post. I will note that you still might find value in reading this if you solved the problem, as my proposed solution also maintains at every stage the graph being a forest (i.e I don't create any cycles).

The Solution

Note that in this editorial, when I write simple path, I mean a simple path between two nodes without having extra edges going from any of the nodes that do not belong to that simple path, or simply a node not connected to anything (this will be clearer once we see drawings).

We should first note that bamboo is simply a tree that looks like a single line of connections (i.e a simple path).

We observe that given the sequence of moves the order we delete/link edges in does not change the resultant underlying graph. We consider the following question: how can we delete edges such that we can link the remaining forest via link operations to make a bamboo after we delete all of them?

Well, we want to achieve a simple path in the end, and a simple path can only be broken into nonintersecting simple paths, therefore after edge deletions, we should have a set of nonintersecting simple paths.

Now, our motivation is to take some pen & paper and draw how such a set looks like (not necessarily the optimal one) (in this editorial I cheat with CSAcademy Graph Editor). **Note that for convenience and as in many tree problems, we root the tree at an arbitrary node.

Suppose we are given the following tree (rooted at node 1):

We can convert it to a set of non-intersecting simple paths in multiple ways, one of which is deleting 4->10, 5->13 & 5->7, and if we were to apply this way we would end up with the following set of simple paths:

If we look at subtrees of the original tree and what they turned into after the deletion, we observe that if a node has outgoing edges to 2 of its original children from the original subtree then that node must not be connected to its parent.

This leads to an approach of dynamic programming over subtrees & an observation is that we can utilize the following state: whether or not our subtree root is connected to its parent. The motivation for this state is exactly what we state above, in order to connect to more than 1 child we must not be connected to our parent.

Let $$$sdp[c]$$$ be the minimum cost to convert $$$c$$$'s subtree into a set of disjoint simple paths, given that we delete the edge from $$$c$$$ to its parent, and let $$$cdp[c]$$$ denote the cost given that we do not delete the edge from node $$$c$$$ to its parent. So in $$$sdp$$$ we allow connecting to 2 children and in $$$cdp$$$ we do not allow that, but for $$$sdp$$$ we add a cost of 1 (deleting the edge to the parent). Note that in this way, the root $$$sdp$$$ 's value will contain 1 more than the actual answer (because the root doesn't have a parent so we add 1 for no reason) but as you'll see we don't use that numerical value in our construction.

For the construction, when we calculate the $$$dp$$$ values we also the children that we keep edges to. let $$$cn$$$ be an array that will store the connections for the $$$cdp$$$. We will assign $$$-1$$$ to $$$cn[c]$$$ if we don't connect anything, and otherwise, we will assign the label of the node that we connect to (note that in this $$$dp$$$ we can connect to almost 1 child because we are already connected to our parent). Furthermore, let $$$sn$$$ be an array of pairs such that it will store the connections for the $$$sdp$$$. We will assign $$$-1$$$ to $$$sn[c].first$$$ if we make no connections. otherwise, we will assign one of the connections to it. Moreover, we will assign $$$-1$$$ to $$$sn[c].second$$$ if we make less than 2 connections, otherwise we will assign to it the label of the node we connect to that we did not assign to $$$sn[c].first$$$.

By using $$$cn$$$ & $$$sn$$$ we will be able to know given the state (connected or not connected to parent) what edges we chose to keep (i.e, the connections we still have).

Now, we shall discuss calculating the $$$dp$$$ values.