WOOHOO, I DID IT!!! sorry, it just took me that long.

And because I'm still high on success, I will leave some tips for despairing ones. (I also don't understand a thing in the editorial)

Calculating every node-combination in a set is O(n²), so we have to condense the set somehow...

Let's say, we have two nodes, v1 and v2; v is the lowest common ancestor(LCA):

      v
     / \
   ..   ..
   /      \
 v1       v2

The thing we have to calculate is v1*v2*d(v1,v2). d() is distance.

It can be split up like this: v1*d(v1,v)*v2 + v2*d(v2,v)*v1

Now, let's add another node (u is the new LCA):

      u
     /  \
    ..   ..
   / \     \
  ..  v2    v3
 /
v1


v1*v2*d(v1,v2) +
(v1*d(v1,u)+v2*d(v2,u))*v3 + v3*d(v3,u)*(v1+v2)

= v1*v2*d(v1,v2) +
((v1+v2)*d(v2,u)+v1*(depth(v1)-depth(v2)))*v3 + v3*d(v3,u)*(v1+v2)

If v1 and v2 are already processed, the only variables are u and v3. We can replace the rest with constants:

C0 + (C1*d(v2,u)+C2)*v3 + v3*d(u,v3)*C1

...there, it's no longer necessay to handle v1 and v2 seperately. (as long as nothing relevant is below v1 or between v1 and v2)

I also used lookup-tables for depth and predecessor.

Please let me know if something is too confusing.