Help_Needed_'s blog

By Help_Needed_, history, 6 weeks ago, In English

Can anyone please help me in one problem?

Problem: You are given a an undirected connected graph, and Q queries, in each query you will be given a vertex 'v' and output for each query should be largest connected component size after removal of vertex 'v'.

number of nodes <= 10^5

queries <= 10^5

Note: Queries are independent.

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6 weeks ago, # |
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Root the tree arbitrarily. Let sz[v] be the size of the subtree of node v, including it (to make the implementation easier).

The answer for query v is max(sz[c] for all children c of v, n-sz[v]), because cutting off node v gives you a tree for each child and another for the rest of nodes. And you can precompute this to not get $$$O(NQ)$$$ runtime.

Total complexity: $$$O(N + Q)$$$

If nodes get removed after each query, then I think you'd need Euler Tour Technique + Segment Tree w/ Lazy Prop to subtract from all nodes in the subtree of v, and maybe something else for the ancestors of v, which would be a lot more complicated. I'm not too sure about this though, so maybe someone knows a better (and simpler!) way of doing this.

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    6 weeks ago, # ^ |
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    If nodes get removed after each query, maybe you could just solve the queries in reverse, and add nodes instead of removing them?

6 weeks ago, # |
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Approach 1. Compute the low function on the graph's dfs tree ( similar to this). When removing node V, the subtree of V's child will be disconnected from the rest of the graph if low[child] > depth[V]. The largest connected component will be equal to Max(disconnected_subtree_sizes, N-sum_of_disconnected_subtree_sizes). You can compute the answer for each node offline, and answer in O(1) for a total of O(n+m).

Approach 2. You can solve the problem using dynamic connectivity ( in O((n+m)log(m)log(n)) by simulating removing all neighbouring edges for each node at a time while also keeping a disjoint set union data structure.