Explanation

The number of severed visits can be computed at the same time while searching for articulation points. There are two cases to consider for node $u$:

1. Node $u$ is not an articulation point, then the number of severed visits by removing the noed is $2 \times (n - 1)$
2. Node $u$ is an articulation point, then the number of severed visits by removing the node is $2 \times (n - 1)$ plus the number of visits whose path goes through node $u$.

Implementation

Time Complexity: $\mathcal{O}(N)$

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#include <bits/stdc++.h>
using namespace std;

int main() {
	int n, m;
	cin >> n >> m;
	vector<vector<int>> g(n + 1);
	for (int i = 0; i < m; i++) {
		int x, y;
		cin >> x >> y;