Open In App
Related Articles

Strongly Connected Components

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report

A strongly connected component is the component of a directed graph that has a path from every vertex to every other vertex in that component. It can only be used in a directed graph.

For example, The below graph has two strongly connected components {1,2,3,4} and {5,6,7} since there is path from each vertex to every other vertex in the same strongly connected component. 

scc_fianldrawio

Strongly Connected Component

Difference between Connected and Strongly Connected Components?

Connectivity in a undirected graph refers to whether two vertices are reachable from each other or not. Two vertices are said to be connected if there is path between them. Meanwhile Strongly Connected is applicable only to directed graphs. A subgraph of a directed graph is considered to be an Strongly Connected Components (SCC) if and only if for every pair of vertices A and B, there exists a path from A to B and a path from B to A. Let’s see why the standard dfs method to find connnected components in a graph cannot be used to determine strongly connected components.

Now, Consider the following graph

scc_fianldrawio

Now, let’s start a dfs call from vertex 1 to visit other vertices.

dfs_finaldrawio

Why conventional DFS method cannot be used to find the strongly connected components?

All the vertices can be reached from vertex 1. But vertices 1 and 5,6,7 can not be in the same strongly connected component because there is no directed path from vertex 5,6 or 7 to vertex 1. The graph has two strongly connected components {1,2,3,4} and {5,6,7}. So the conventional dfs method cannot be used to find the strongly connected components.

Connecting Two Strongly Connected Component by a Unidirectional Edge

Two different connected components becomes a single component if a edge is added between a vertex from one component to a vertex of other component. But this is not the case in strongly connected components. Two strongly connected components doesn’t become a single strongly connected component if there is only a unidirectional edge from one SCC to other SCC.

unidrawio-(2)


Brute Force Approach for Finding Strongly Connected Components:

The simple method will be for each vertex i (which is not a part of any strongly component) find the vertices which will be the part of strongly connected component containing vertex i. Two vertex i and j will be in the same strongly connected component if they there is a directed path from vertex i to vertex j and vice-versa.

Let’s understand the approach with the help of following example:-

exampledrawio

  • Starting with vertex 1. There is path from vertex 1 to vertex 2 and vice-versa. Similarly there is a path from vertex 1 to vertex 3 and vice versa. So, vertex 2 and 3 will be in the same Strongly Connected Component as vertex 1. Although there is directed path form vertex 1 to vertex 4 and vertex 5. But there is no directed path from vertex 4,5 to vertex 1 so vertex 4 and 5 won’t be in the same Strongly Connected Component as vertex 1. Thus Vertex 1,2 and 3 forms a Strongly Connected Component.
  • For Vertex 2 and 3, there Strongly Connected Component has already been determined.
  • For Vertex 4, there is a path from vertex 4 to vertex 5 but there is no path from vertex 5 to vertex 4. So vertex 4 and vertex won’t be in the Same Strongly Connected Component. So both Vertex 4 and Vertex 5 will be part of Single Strongly Connected Component.
  • Hence there will be 3 Strongly Connected Component {1,2,3}, {4} and {5}.

Below is the implementation of above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
class GFG {
public:
    // dfs Function to reach destination
    bool dfs(int curr, int des, vector<vector<int> >& adj,
             vector<int>& vis)
    {
 
        // If curr node is destination return true
        if (curr == des) {
            return true;
        }
        vis[curr] = 1;
        for (auto x : adj[curr]) {
            if (!vis[x]) {
                if (dfs(x, des, adj, vis)) {
                    return true;
                }
            }
        }
        return false;
    }
 
    // To tell whether there is path from source to
    // destination
    bool isPath(int src, int des, vector<vector<int> >& adj)
    {
        vector<int> vis(adj.size() + 1, 0);
        return dfs(src, des, adj, vis);
    }
 
    // Function to return all the strongly connected
    // component of a graph.
    vector<vector<int> > findSCC(int n,
                                 vector<vector<int> >& a)
    {
        // Stores all the strongly connected components.
        vector<vector<int> > ans;
 
        // Stores whether a vertex is a part of any Strongly
        // Connected Component
        vector<int> is_scc(n + 1, 0);
 
        vector<vector<int> > adj(n + 1);
 
        for (int i = 0; i < a.size(); i++) {
            adj[a[i][0]].push_back(a[i][1]);
        }
 
        // Traversing all the vertices
        for (int i = 1; i <= n; i++) {
 
            if (!is_scc[i]) {
 
                // If a vertex i is not a part of any SCC
                // insert it into a new SCC list and check
                // for other vertices whether they can be
                // thr part of thidl ist.
                vector<int> scc;
                scc.push_back(i);
 
                for (int j = i + 1; j <= n; j++) {
 
                    // If there is a path from vertex i to
                    // vertex j and vice versa put vertex j
                    // into the current SCC list.
                    if (!is_scc[j] && isPath(i, j, adj)
                        && isPath(j, i, adj)) {
                        is_scc[j] = 1;
                        scc.push_back(j);
                    }
                }
 
                // Insert the SCC containing vertex i into
                // the final list.
                ans.push_back(scc);
            }
        }
        return ans;
    }
};
 
// Driver Code Starts
 
int main()
{
 
    GFG obj;
    int V = 5;
    vector<vector<int> > edges{
        { 1, 3 }, { 1, 4 }, { 2, 1 }, { 3, 2 }, { 4, 5 }
    };
    vector<vector<int> > ans = obj.findSCC(V, edges);
    cout << "Strongly Connected Components are:\n";
    for (auto x : ans) {
        for (auto y : x) {
            cout << y << " ";
        }
        cout << "\n";
    }
}

                    

Java

import java.util.ArrayList;
import java.util.List;
 
class GFG {
 
    // dfs Function to reach destination
    boolean dfs(int curr, int des, List<List<Integer>> adj,
                List<Integer> vis) {
 
        // If curr node is destination return true
        if (curr == des) {
            return true;
        }
        vis.set(curr, 1);
        for (int x : adj.get(curr)) {
            if (vis.get(x) == 0) {
                if (dfs(x, des, adj, vis)) {
                    return true;
                }
            }
        }
        return false;
    }
 
    // To tell whether there is path from source to
    // destination
    boolean isPath(int src, int des, List<List<Integer>> adj) {
        List<Integer> vis = new ArrayList<>(adj.size() + 1);
        for (int i = 0; i <= adj.size(); i++) {
            vis.add(0);
        }
        return dfs(src, des, adj, vis);
    }
 
    // Function to return all the strongly connected
    // component of a graph.
    List<List<Integer>> findSCC(int n, List<List<Integer>> a) {
        // Stores all the strongly connected components.
        List<List<Integer>> ans = new ArrayList<>();
 
        // Stores whether a vertex is a part of any Strongly
        // Connected Component
        List<Integer> is_scc = new ArrayList<>(n + 1);
        for (int i = 0; i <= n; i++) {
            is_scc.add(0);
        }
 
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            adj.add(new ArrayList<>());
        }
 
        for (List<Integer> edge : a) {
            adj.get(edge.get(0)).add(edge.get(1));
        }
 
        // Traversing all the vertices
        for (int i = 1; i <= n; i++) {
 
            if (is_scc.get(i) == 0) {
 
                // If a vertex i is not a part of any SCC
                // insert it into a new SCC list and check
                // for other vertices whether they can be
                // the part of this list.
                List<Integer> scc = new ArrayList<>();
                scc.add(i);
 
                for (int j = i + 1; j <= n; j++) {
 
                    // If there is a path from vertex i to
                    // vertex j and vice versa, put vertex j
                    // into the current SCC list.
                    if (is_scc.get(j) == 0 && isPath(i, j, adj)
                            && isPath(j, i, adj)) {
                        is_scc.set(j, 1);
                        scc.add(j);
                    }
                }
 
                // Insert the SCC containing vertex i into
                // the final list.
                ans.add(scc);
            }
        }
        return ans;
    }
}
 
public class Main {
 
    public static void main(String[] args) {
 
        GFG obj = new GFG();
        int V = 5;
        List<List<Integer>> edges = new ArrayList<>();
        edges.add(new ArrayList<>(List.of(1, 3)));
        edges.add(new ArrayList<>(List.of(1, 4)));
        edges.add(new ArrayList<>(List.of(2, 1)));
        edges.add(new ArrayList<>(List.of(3, 2)));
        edges.add(new ArrayList<>(List.of(4, 5)));
        List<List<Integer>> ans = obj.findSCC(V, edges);
        System.out.println("Strongly Connected Components are:");
        for (List<Integer> x : ans) {
            for (int y : x) {
                System.out.print(y + " ");
            }
            System.out.println();
        }
    }
}
 
// This code is contributed by shivamgupta310570

                    

Python3

class GFG:
    # dfs Function to reach destination
    def dfs(self, curr, des, adj, vis):
        # If current node is the destination, return True
        if curr == des:
            return True
        vis[curr] = 1
        for x in adj[curr]:
            if not vis[x]:
                if self.dfs(x, des, adj, vis):
                    return True
        return False
     
    # To tell whether there is a path from source to destination
    def isPath(self, src, des, adj):
        vis = [0] * (len(adj) + 1)
        return self.dfs(src, des, adj, vis)
     
    # Function to return all the strongly connected components of a graph.
    def findSCC(self, n, a):
        # Stores all the strongly connected components.
        ans = []
         
        # Stores whether a vertex is a part of any Strongly Connected Component
        is_scc = [0] * (n + 1)
         
        adj = [[] for _ in range(n + 1)]
         
        for i in range(len(a)):
            adj[a[i][0]].append(a[i][1])
         
        # Traversing all the vertices
        for i in range(1, n + 1):
            if not is_scc[i]:
                # If a vertex i is not a part of any SCC, insert it into a new SCC list
                # and check for other vertices whether they can be part of this list.
                scc = [i]
                for j in range(i + 1, n + 1):
                    # If there is a path from vertex i to vertex j and vice versa,
                    # put vertex j into the current SCC list.
                    if not is_scc[j] and self.isPath(i, j, adj) and self.isPath(j, i, adj):
                        is_scc[j] = 1
                        scc.append(j)
                # Insert the SCC containing vertex i into the final list.
                ans.append(scc)
        return ans
 
# Driver Code Starts
if __name__ == "__main__":
    obj = GFG()
    V = 5
    edges = [
        [1, 3], [1, 4], [2, 1], [3, 2], [4, 5]
    ]
    ans = obj.findSCC(V, edges)
    print("Strongly Connected Components are:")
    for x in ans:
        for y in x:
            print(y, end=" ")
        print()
         
# This code is contributed by shivamgupta310570

                    

C#

using System;
using System.Collections.Generic;
 
class GFG
{
    // dfs Function to reach destination
    public bool Dfs(int curr, int des, List<List<int>> adj, List<int> vis)
    {
        // If curr node is the destination, return true
        if (curr == des)
        {
            return true;
        }
        vis[curr] = 1;
        foreach (var x in adj[curr])
        {
            if (vis[x] == 0)
            {
                if (Dfs(x, des, adj, vis))
                {
                    return true;
                }
            }
        }
        return false;
    }
 
    // To tell whether there is a path from source to destination
    public bool IsPath(int src, int des, List<List<int>> adj)
    {
        var vis = new List<int>(adj.Count + 1);
        for (int i = 0; i < adj.Count + 1; i++)
        {
            vis.Add(0);
        }
        return Dfs(src, des, adj, vis);
    }
 
    // Function to return all the strongly connected components of a graph
    public List<List<int>> FindSCC(int n, List<List<int>> a)
    {
        // Stores all the strongly connected components
        var ans = new List<List<int>>();
 
        // Stores whether a vertex is a part of any Strongly Connected Component
        var isScc = new List<int>(n + 1);
        for (int i = 0; i < n + 1; i++)
        {
            isScc.Add(0);
        }
 
        var adj = new List<List<int>>(n + 1);
        for (int i = 0; i < n + 1; i++)
        {
            adj.Add(new List<int>());
        }
 
        for (int i = 0; i < a.Count; i++)
        {
            adj[a[i][0]].Add(a[i][1]);
        }
 
        // Traversing all the vertices
        for (int i = 1; i <= n; i++)
        {
            if (isScc[i] == 0)
            {
                // If a vertex i is not a part of any SCC
                // insert it into a new SCC list and check
                // for other vertices whether they can be
                // the part of this list.
                var scc = new List<int>();
                scc.Add(i);
 
                for (int j = i + 1; j <= n; j++)
                {
                    // If there is a path from vertex i to
                    // vertex j and vice versa, put vertex j
                    // into the current SCC list.
                    if (isScc[j] == 0 && IsPath(i, j, adj) && IsPath(j, i, adj))
                    {
                        isScc[j] = 1;
                        scc.Add(j);
                    }
                }
 
                // Insert the SCC containing vertex i into
                // the final list.
                ans.Add(scc);
            }
        }
        return ans;
    }
}
 
// Driver Code Starts
class Program
{
    static void Main(string[] args)
    {
        GFG obj = new GFG();
        int V = 5;
        List<List<int>> edges = new List<List<int>>
        {
            new List<int> { 1, 3 }, new List<int> { 1, 4 }, new List<int> { 2, 1 },
            new List<int> { 3, 2 }, new List<int> { 4, 5 }
        };
        List<List<int>> ans = obj.FindSCC(V, edges);
        Console.WriteLine("Strongly Connected Components are:");
        foreach (var x in ans)
        {
            foreach (var y in x)
            {
                Console.Write(y + " ");
            }
            Console.WriteLine();
        }
    }
}
 
 
// This code is contributed by shivamgupta310570

                    

Javascript

class GFG {
    // Function to reach the destination using DFS
    dfs(curr, des, adj, vis) {
        // If the current node is the destination, return true
        if (curr === des) {
            return true;
        }
        vis[curr] = 1;
        for (let x of adj[curr]) {
            if (!vis[x]) {
                if (this.dfs(x, des, adj, vis)) {
                    return true;
                }
            }
        }
        return false;
    }
 
    // Check whether there is a path from source to destination
    isPath(src, des, adj) {
        const vis = new Array(adj.length + 1).fill(0);
        return this.dfs(src, des, adj, vis);
    }
 
    // Function to find all strongly connected components of a graph
    findSCC(n, a) {
        // Stores all strongly connected components
        const ans = [];
 
        // Stores whether a vertex is part of any Strongly Connected Component
        const is_scc = new Array(n + 1).fill(0);
        const adj = new Array(n + 1).fill().map(() => []);
 
        for (let i = 0; i < a.length; i++) {
            adj[a[i][0]].push(a[i][1]);
        }
 
        // Traversing all the vertices
        for (let i = 1; i <= n; i++) {
            if (!is_scc[i]) {
                // If a vertex i is not part of any SCC,
                // insert it into a new SCC list and check
                // for other vertices that can be part of this list.
                const scc = [i];
                for (let j = i + 1; j <= n; j++) {
                    // If there is a path from vertex i to
                    // vertex j and vice versa, put vertex j
                    // into the current SCC list.
                    if (!is_scc[j] && this.isPath(i, j, adj) && this.isPath(j, i, adj)) {
                        is_scc[j] = 1;
                        scc.push(j);
                    }
                }
                // Insert the SCC containing vertex i into the final list.
                ans.push(scc);
            }
        }
        return ans;
    }
}
 
// Driver Code Starts
const obj = new GFG();
const V = 5;
const edges = [
    [1, 3], [1, 4], [2, 1], [3, 2], [4, 5]
];
const ans = obj.findSCC(V, edges);
console.log("Strongly Connected Components are:");
for (let x of ans) {
    console.log(x.join(" "));
}
 
// This code is contributed by shivamgupta310570

                    

Output
Strongly Connected Components are:
1 2 3 
4 
5 









Time complexity: O(n * (n + m)), because for each pair of vertices we are checking whether a path exists between them.
Auxiliary Space: O(N) 

Related article for finding strongly connected components in a directed graph in O(V+E) time:



Last Updated : 17 Jan, 2024
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads