So in this problem 1294F - Три пути в дереве. The editorial solution is this :
Spoiler
Suppose our tree has 23 nodes. Let the diameter contains all the node b/w 1 and 20 (take in increasing order for simplicity). Lets take a branch from node 5 which contains the remaining nodes . So according to the editorial we use multi-source bfs and find the node which is at farthest distance . The node that will be farthest will be around node 9 or 10. But if we take this as third node the result will be (1 , 20 , (9 or 10)) but this will not increase the result we will still get 19 edges as our answer. Rather we should take a branch with is at max distance from any node on diameter. I tried this approach but it is giving MLE . Here is the code:
