Minimize pairwise connectivity after removal of a node

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~~Can anyone please help me in one p~~Problem?:↵
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Problem: You are given an undirected connected graph, and a set of Q queries~~, in e~~. Each query ~~you will be given a vertex 'v' and output for each query should be v(v-1)/2 with remains component a~~involves removing a vertex 'v' from the graph. After remov~~al of~~ing vertex 'v'~~. When removing node v~~, the subtree of 'v's child ~~will be~~is disconnected from the rest of the graph and ~~also separate from the main graph. Each of the part of the graph~~is treated as a separate component. For each such component, calculate the value v(v-1)/2, where v is the count of vertices in that component.↵
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The goal is to minimize the pairwise connectivity after removing ~~one~~each node perform this v(v-1)/2 where v is the count of the total connected component this part.↵
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Now you have to minimize the pairwise connectivity after removing each node.↵
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nodes <= 20^5; queries <= 20^5 (Queries are independent). The pairwise connectivity is the sum of v(v-1)/2 for all components resulting from the removal of the node 'v'.↵
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Input:↵
A connected undirected graph with nodes (vertices) and edges.↵
Q queries, each specifying a vertex 'v' to be removed.↵
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Output:↵
For each query, output the minimum pairwise connectivity after removing the specified vertex 'v' according to the rules defined above.↵
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Constraints:↵
The number of nodes (vertices) in the graph is less than or equal to 20^5.↵
The number of queries is less than or equal to 20^5.↵
Queries are independent of each other.↵
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Example:↵
Input:↵
- Graph: {1-2, 1-3, 2-4, 2-5, 3-6}↵
- Queries: Q1(1), Q2(2), Q3(3)↵
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Output:↵
- After removing vertex 1: Pairwise connectivity = 4↵
- After removing vertex 2: Pairwise connectivity = 4↵
- After removing vertex 3: Pairwise connectivity = 2↵
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Note: The provided example is for illustration purposes only and may not represent the actual solution to the problem. The actual solution depends on the specific structure of the given graph and queries.