BFS for Disconnected Graph

In previous post, BFS only with a particular vertex is performed i.e. it is assumed that all vertices are reachable from the starting vertex. But in the case of disconnected graph or any vertex that is unreachable from all vertex, the previous implementation will not give the desired output, so in this post, a modification is done in BFS.


All vertices are reachable. So, for above graph simple BFS will work.

graph
As in above graph a vertex 1 is unreachable from all vertex, so simple BFS wouldn’t work for it.



Just to modify BFS, perform simple BFS from each 
unvisited vertex of given graph.

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// C++ implementation of modified BFS
#include<bits/stdc++.h>
using namespace std;
  
// A utility function to add an edge in an
// undirected graph.
void addEdge(vector<int> adj[], int u, int v)
{
    adj[u].push_back(v);
}
  
// A utility function to do BFS of graph
// from a given vertex u.
void BFSUtil(int u, vector<int> adj[],
            vector<bool> &visited)
{
  
    // Create a queue for BFS
    list<int> q;
   
    // Mark the current node as visited and enqueue it
    visited[u] = true;
    q.push_back(u);
   
    // 'i' will be used to get all adjacent vertices 4
    // of a vertex list<int>::iterator i;
   
    while(!q.empty())
    {
        // Dequeue a vertex from queue and print it
        u = q.front();
        cout << u << " ";
        q.pop_front();
   
        // Get all adjacent vertices of the dequeued
        // vertex s. If an adjacent has not been visited, 
        // then mark it visited and enqueue it
        for (int i = 0; i != adj[u].size(); ++i)
        {
            if (!visited[adj[u][i]])
            {
                visited[adj[u][i]] = true;
                q.push_back(adj[u][i]);
            }
        }
    }
}
  
// This function does BFSUtil() for all 
// unvisited vertices.
void BFS(vector<int> adj[], int V)
{
    vector<bool> visited(V, false);
    for (int u=0; u<V; u++)
        if (visited[u] == false)
            BFSUtil(u, adj, visited);
}
  
// Driver code
int main()
{
    int V = 5;
    vector<int> adj[V];
  
    addEdge(adj, 0, 4);
    addEdge(adj, 1, 2);
    addEdge(adj, 1, 3);
    addEdge(adj, 1, 4);
    addEdge(adj, 2, 3);
    addEdge(adj, 3, 4);
    BFS(adj, V);
    return 0;
}

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Output:

0 4 1 2 3

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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