Critical Connections in a Network
<p>There are <code>n</code> servers numbered from <code>0</code> to <code>n - 1</code> connected by undirected server-to-server <code>connections</code> forming a network where <code>connections[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> represents a connection between servers <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>. Any server can reach other servers directly or indirectly through the network.</p>

<p>A <em>critical connection</em> is a connection that, if removed, will make some servers unable to reach some other server.</p>

<p>Return all critical connections in the network in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1192.Critical%20Connections%20in%20a%20Network/images/1537_ex1_2.png" style="width: 198px; height: 248px;" />
<pre>
<strong>Input:</strong> n = 4, connections = [[0,1],[1,2],[2,0],[1,3]]
<strong>Output:</strong> [[1,3]]
<strong>Explanation:</strong> [[3,1]] is also accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2, connections = [[0,1]]
<strong>Output:</strong> [[0,1]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>n - 1 &lt;= connections.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated connections.</li>
</ul>
Depth-First Search; Graph; Biconnected Component
class Solution {
public:
    vector<vector<int>> criticalConnections(int n, vector<vector<int>>& connections) {
        int now = 0;
        vector<int> dfn(n);
        vector<int> low(n);
        vector<int> g[n];
        for (auto& e : connections) {
            int a = e[0], b = e[1];
            g[a].push_back(b);
            g[b].push_back(a);
        }
        vector<vector<int>> ans;
        function<void(int, int)> tarjan = [&](int a, int fa) -> void {
            dfn[a] = low[a] = ++now;
            for (int b : g[a]) {
                if (b == fa) {
                    continue;
                }
                if (!dfn[b]) {
                    tarjan(b, a);
                    low[a] = min(low[a], low[b]);
                    if (low[b] > dfn[a]) {
                        ans.push_back({a, b});
                    }
                } else {
                    low[a] = min(low[a], dfn[b]);
                }
            }
        };
        tarjan(0, -1);
        return ans;
    }
};