No, There won't be any onsite this time.

TDT is just standard modification of Barricades problem from Looking for a challenge book. You can do dp[x][y] which denotes maximum value of a component in subtree of x containing y nodes including x. Now, it is trivial that complexity to compute this dp is less than equal to O(n^3). But if you iterate only over current size of subtree then it can be computed in O(n^2). You can get it that you will only iterate number of times equal to number of times two nodes's lca is x while computing dp for x. So, it is O(n^2). Sorry if I was unclear. My code- LINK

Firstly, without loss of generality we can assume the tree to be rooted at 1. Let dp[i][j] be the maximum sum(which in turn will lead to maximum mean) that can be obtained by considering the j-connected component in the subtree of node i which includes node i.Now for a node src, if I have the DP table filled for all of its children, how can I fill up the DP table for node src?
Here is the code for your reference. You need to run it for all children u of node src.
Lastly, do not forget to take floor value at the end.