A connected subgraph of $G$ that is not a proper subgraph of any other connected subgraph of $G$ is a **component** of $G$, i.e. there’s a $u-v$ path in the mentioned subgraph

## Undirected graphs

The problem of finding components in an undirected graph requires a simple graph traversal starting from an arbitrary vertex keeping track of the vertices that were already visited, it’s also needed to run the algorithm above for every vertex of $G$ (given that it was not visited)

- the number of components of an undirected graph $G$ is equal to the number of disconnected subgraphs

```
vector<bool> visited;
// adjacency list of G
vector<vector<int> > g;
void dfs(int v) {
visited[v] = true;
for (int i = 0; i < g[v].size(); i += 1) {
int next = g[v][i];
if (!visited[next]) {
dfs(next);
}
}
}
/**
* Computes the number of connected components in an undirected graph `G`
* of order `n` and size `m`
*
* Time complexity: O(n + m)
* Space complexity: O(n)
*
* @return {int} The number of components in `G`
*/
int connected_components() {
int n = g.size();
visited.assign(n, false);
int components = 0;
for (int i = 0; i < visited.size(); i += 1) {
if (!visited[i]) {
dfs(i);
++components;
}
}
return components;
}
```

## Directed graphs

Given a directed graph $G$ two nodes $u, v \in V(G)$ are called **strongly connected** if $v$ is reachable from $u$ and $u$ is reachable from $v$

A **strongly connected component** (SCC) of $G$ is a subgraph $C \subseteq V(G)$ such that

- $C$ is not empty
- for any $u,v \in V(G)$, $u$ and $v$ are strongly connected
- for any $u \in V(G)$ and $v \in G - C$, $u$ and $v$ are not strongly connected

### Tarjan’s algorithm

The idea is to perform a DFS from an arbitrary vertex (conducting subsequent DFS from non-explored vertices), during the traversal each vertex $v$ is assigned with two numbers:

- the
**time**it was explored denoted as $v_{time}$ - the smallest index of any node known to be reachable from $v$ denoted as $v_{low}$

Let $u$ be a node that belongs to a SCC, if $v$ is the arbitrary vertex chosen then the only known vertex that is reachable from $u$ is $u$, let $v$ be a vertex discovered during the exploration of $u$, if there’s a $v \rightarrow u$ path then it means that there’s a cycle and all the vertices in the path $u-v$ belong to the same connected component, such a node $u$ is called the **root of the SCC**

Let $u$ be a node that belongs to a SCC, if it’s known that there’s a $u-v$ cycle and also that $u$ can reach a vertex $t$ with lower index than $u$ then $v$ and $t$ belong to the same component

A stack is also needed to keep track of the nodes that were visited, the working of the stack follows the invariant: a node remains on the stack after exploration if and only if it has a path to some node earlier in the stack

```
// adjacency list of G
vector<vector<int> > g;
int time_spent;
// the number of scc
int total_scc;
// the time a vertex was discovered
vector<int> time_in;
// the smallest index of any vertex known to be reachable from `i`
vector<int> back;
// the scc vertex `i` belongs to
vector<int> scc;
// invariant: a node remains in the stack after exploration if
// it has a path to some node explored earlier that is in the stack
vector<bool> in_stack;
stack<int> vertices;
void dfs(int v) {
int next;
// the lowest back edge discovery time of `v` is
// set to the discovery time of `v` initally
back[v] = time_in[v] = ++time_spent;
vertices.push(v);
in_stack[v] = true;
for (int i = 0; i < g[v].size(); i += 1) {
next = g[v][i];
if (time_in[next] == -1) {
// unvisited edge
dfs(next);
// propagation of the lowest back edge discovery time
back[v] = min(back[v], back[next]);
} else if (in_stack[next]) {
// (v, next) is a back edge only if it's connected to a predecessor
// of `v`, i.e. if `next` is in same branch in the dfs tree
//
// an alternative is to use the time a vertex finished exploring its
// adjacent nodes, if the time is not set then it's a back edge
back[v] = min(back[v], time_in[next]);
}
}
// if the root node of a connected component has finished
// exploring all its neighbors, assign the same component `id`
// to all the elements in the scc
if (back[v] == time_in[v]) {
total_scc += 1;
do {
next = vertices.top();
vertices.pop();
in_stack[next] = false;
scc[next] = total_scc;
} while (next != v);
}
}
/**
* Finds the strongly connected components in a digraph `G` of order `n`
* and size `m`
*
* Time complexity: O(n + m)
* Space complexity: O(n)
*
* @returns {int} the number of strongly connected components
*/
int tarjan() {
int n = g.size();
scc.assign(n, -1);
time_in.assign(n, -1);
back.assign(n, -1);
in_stack.assign(n, false);
while (!vertices.empty()) {
vertices.pop();
}
time_spent = 0;
total_scc = 0;
for (int i = 0; i < n; i += 1) {
if (time_in[i] == -1) {
dfs(i);
}
}
return total_scc;
}
```