Given an undirected graph, print all connected components line by line. For example consider the following graph.

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We have discussed algorithms for finding strongly connected components in directed graphs in following posts.

Kosaraju’s algorithm for strongly connected components.

Tarjan’s Algorithm to find Strongly Connected Components

Finding connected components for an undirected graph is an easier task. We simple need to do either BFS or DFS starting from every unvisited vertex, and we get all strongly connected components. Below are steps based on DFS.

1) Initialize all vertices as not visited. 2) Do following for every vertex 'v'. (a) If 'v' is not visited before, call DFSUtil(v) (b) Print new line character DFSUtil(v) 1) Mark 'v' as visited. 2) Print 'v' 3) Do following for every adjacent 'u' of 'v'. If 'u' is not visited, then recursively call DFSUtil(u)

Below is C++ implementation of above algorithm.

// C++ program to print connected components in // an undirected graph #include<iostream> #include <list> using namespace std; // Graph class represents a undirected graph // using adjacency list representation class Graph { int V; // No. of vertices // Pointer to an array containing adjacency lists list<int> *adj; // A function used by DFS void DFSUtil(int v, bool visited[]); public: Graph(int V); // Constructor void addEdge(int v, int w); void connectedComponents(); }; // Method to print connected components in an // undirected graph void Graph::connectedComponents() { // Mark all the vertices as not visited bool *visited = new bool[V]; for(int v = 0; v < V; v++) visited[v] = false; for (int v=0; v<V; v++) { if (visited[v] == false) { // print all reachable vertices // from v DFSUtil(v, visited); cout << "\n"; } } } void Graph::DFSUtil(int v, bool visited[]) { // Mark the current node as visited and print it visited[v] = true; cout << v << " "; // Recur for all the vertices // adjacent to this vertex list<int>::iterator i; for(i = adj[v].begin(); i != adj[v].end(); ++i) if(!visited[*i]) DFSUtil(*i, visited); } Graph::Graph(int V) { this->V = V; adj = new list<int>[V]; } // method to add an undirected edge void Graph::addEdge(int v, int w) { adj[v].push_back(w); adj[w].push_back(v); } // Drive program to test above int main() { // Create a graph given in the above diagram Graph g(5); // 5 vertices numbered from 0 to 4 g.addEdge(1, 0); g.addEdge(2, 3); g.addEdge(3, 4); cout << "Following are connected components \n"; g.connectedComponents(); return 0; }

Output

0 1 2 3 4

Time complexity of above solution is O(V + E) as it does simple DFS for given graph.

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