(Analysis by Travis Hance)

Since $K$ is very small and $N$ can be very large, this suggests trying to find a solution which is linear in $N$ and possibly exponential in $K$. This suggests a dynamic programming approach where we proceed one row at a time, building a minimum spanning tree as we go. In general, this partially constructed MST will be a forest, and we must keep track of which nodes are in the same component of the forest. If two nodes are in the same component, we cannot add a new edge which connects them; conversely, if two nodes are not in the same component, then they must eventually be connected.

For our first attempt, our DP state can be $(i,r)$ where $1 \le i \le N$ is the row that we have built our forest up to, and $r$ is a state which represents which of the $K$ nodes in this row are connected to each other by the partially constructed tree. For each state, the value we will track will be the total weight of the minimum forest in the first $i$ rows such that:

And of course, we also track the total number of such forests with that exact weight.

One way to represent $r$ is by a sequence of $k$ integers, where each integer represents a component. For example, $[1,2,3,4,5,6]$ represents a row where all nodes are in different components, and $[1,1,1,1,1,1]$ represents a row where each node is connected. After accounting for many of these states being equivalent, and accounting for the fact that some, like $[1,2,1,2]$ are impossible (because the graph must be planar), we count a total of $132$ states for $K=6$. It is easiest to pre-compute all of these states before beginning the DP. In general, the number of states are gives by the Catalan sequence $C_K$.

To transition from one row to the other, we have to consider every possible subset of edges between the two rows that we could include in the forest, then consider how the state is transformed by those edges. This gives $2^{2K-1}$ possible ways to transition any state (there are $N$ edges between two rows and $N-1$ edges in the next row). This row-by-row approach gives a runtime $O(NKC_K 2^{2K-1})$.

We can do better by not trying to transition the entire row at once. In fact, this approach is not just quicker, but it is likely simpler, simply because there are only a constant number of transitions fore each state. Our state is now $(i, j, r)$, which now corresponds to the first $i$ rows plus the leftmost $j$ nodes of row $i+1$. There are two ways to transition from $(i, K, r)$ to $(i+1, 1, *)$ (there is one edge, which can choose to take or not), and there are four ways to transition from $(i, j, r)$ to $(i, j+1, *)$ (there are two edges to choose from). This gives a runtime of $O(NKC_K)$.