Hello everyone. I participated in the lowe's coding round and there was this question I was not able to solve completely. I did brute force and passed some test cases though.

The problem says, You are given a connected undirected graph. It has n nodes and m edges. Each node has some sweetness value that will be given to us.

The game is as follows:

- Alice moves first and she may break a node. Each edge connected to this node will vanish.
- Bob will pick any connected component(containing all or some nodes)
- Alice will pick any remaining connected components if there are any

The game ends in three steps. Both of them want to maximize their score by collecting maximum possible sweetness. They are not trying to minimize each other' score.

Determine the maximum score of Alice and Bob respectively. Assume both plays the game optimally.

Graph nodes index starts from 1. If no connected component left, alice gets 0

Input: number of test cases then n, m and then sweetness of each node and then the edges.

Example:

1

6 7

4 3 7 5 9 2

1 2

2 3

1 3

3 4

4 5

5 6

4 6

For this answer is 11 14

I tried to construct a spanning tree and also strongly connected components but couldn't understand.

How to solve this problem? I did this with DFS and max heap. Brute force. What is optimal way of doing this problem

Lowe's hiring challenge is still ongoing till midnight today. So, please ask tomorrow.

Round 1 as well? This one came in 1st round.

Then, you should write in

BOLDthat this was from round 1. Since, round 2 of the same`Lowe's hiring coding contest`

was held today and you wrote this blog today, so anyone's intuition will say that it is from today's round. In fact, there were many guys who did the same thing.These hiring challenges are really destroying competitive programming.

Can you please elaborate why you think so?

It has created a lot of desperate people who do not actually care about competitive programming but need to be good anyway. This in turn has increased the number of people with a shitty attitude, including but not limited to cheaters.

I do agree, these type of hirings have cultivated the cheater culture.

Sir the contest is over I am not doing cheating in any way. Can't I ask a simple question on CF?

In a good way or you mean sarcastically :)

Can you explain exactly why you think so ? Cuz, the companies can't interview all the candidates who apply for a particular job profile, as there are too many of them. Also, they can't give standard interview problems which can be googled easily. So, they have to stick to something similar to competitive programming. Although, the problems are much easier and many times are slight variation of the standard algorithms.

You have explained why these companies do this, but this is not at all a refutation of "hiring challenges are destroying competitive programming".

Yes, but my point was the companies "have" to do it. I agree with your point though. Moreover, many people (not all) who get hired by these challenges often start a youtube channel where they propagate many misconceptions regarding competitive programming, resulting in more cheating.

round 1 is over

Were you able to get AC on this one?

sorry , but I had uploaded the black node solution, i thought you were asking black node

The solution is to use the DFS tree.

Can you please elaborate further because my dfs gave me TLE. I deleted all nodes and then computed cost every time.

In the following, I'll denote as $$$\mathrm{sw}[u]$$$ the sweetness value of $$$u$$$. Build the DFS tree (shamelessly linking to my own blog, oh no) of the graph. In every vertex $$$v$$$, store the total sweetness value of the subtree of that vertex, call that value $$$\mathrm{sub}[v]$$$.

Let's consider removing some vertex $$$u$$$. What connected components are there after removing the vertex $$$u$$$? For every child $$$v$$$ of $$$u$$$

such that there is no back-edge from a descendant of $$$v$$$ to a proper ancestor of $$$u$$$, there will be a component with total sweetness value $$$\mathrm{sub}[v]$$$. The rest of the vertices will be in one connected component connected to the parent of $$$u$$$. We have found the sizes of all these components, so we can calculate the value of the largest one, give it to Bob and give all other vertices to Alice. Thus, we can check the scores of both players if Alice removes the vertex $$$u$$$ at first. Calculate these scores for every choice of $$$u$$$.The complexity will be $$$O(m)$$$.