#include  using namespace std; class GFG { public:  // dfs Function to reach destination  bool dfs(int curr, int des, vector>& agg, vettore & vis) { // Se il nodo curr è la destinazione 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;  } } } restituisce falso;  } // Per sapere se esiste un percorso dall'origine alla // destinazione bool isPath(int src, int des, vettore>& agg) { vettore vis(agg.dimensione() + 1, 0);  return dfs(src, des, adj, vis);  } // Funzione per restituire tutti i // componenti fortemente connessi di un grafico.  vettore> findSCC(int n, vettore>& a) { // Memorizza tutti i componenti fortemente connessi.  vettore> ans;  // Memorizza se un vertice fa parte di qualsiasi vettore componente fortemente // connesso è_scc(n + 1, 0);  vettore> agg(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 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> bordi{ { 1, 3 }, { 1, 4 }, { 2, 1 }, { 3, 2 }, { 4, 5 } };  vettore> ans = obj.findSCC(V, bordi);  cout<< 'Strongly Connected Components are:
';  for (auto x : ans) {  for (auto y : x) {  cout << y << ' ';  }  cout << '
';  } }>

import java.util.ArrayList; import java.util.List; class GFG {  // dfs Function to reach destination  boolean dfs(int curr, int des, List> agg., Lista vis) { // Se il nodo curr è la destinazione 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;  } } } restituisce falso;  } // Per sapere se esiste un percorso dall'origine alla // destinazione boolean isPath(int src, int des, List> agg) { Lista 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> findSCC(int n, List> a) { // Memorizza tutti i componenti fortemente connessi.  Elenco> ans = new ArrayList();  // Memorizza se un vertice fa parte di un elenco di componenti fortemente // connessi is_scc = nuovo ArrayList(n + 1);  for (int i = 0; i<= n; i++) {  is_scc.add(0);  }  List> adj = nuovo ArrayList();  for (int i = 0; i<= n; i++) {  adj.add(new ArrayList());  }  for (List bordo : a) { adj.get(edge.get(0)).add(edge.get(1));  } // Attraversa tutti i vertici per (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 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> bordi = new ArrayList();  bordi.add(new ArrayList(List.of(1, 3)));  bordi.add(new ArrayList(List.of(1, 4)));  bordi.add(new ArrayList(List.of(2, 1)));  bordi.add(new ArrayList(List.of(3, 2)));  bordi.add(new ArrayList(List.of(4, 5)));  Elenco> ans = obj.findSCC(V, bordi);  System.out.println('I componenti fortemente connessi sono:');  per (Elenco x : ans) { for (int y : x) { System.out.print(y + ' ');  } System.out.println();  } } } // Questo codice è stato fornito da shivamgupta310570>

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>

using System; using System.Collections.Generic; class GFG {  // dfs Function to reach destination  public bool Dfs(int curr, int des, List> agg., Lista vis) { // Se il nodo curr è la destinazione, restituisce 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;  } } } restituisce falso;  } // Per sapere se esiste un percorso dall'origine alla destinazione public bool IsPath(int src, int des, List> agg) { var show = nuova lista (Conteggio agg. + 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> FindSCC(int n, List> a) { // Memorizza tutti i componenti fortemente connessi var ans = new List>();  // Memorizza se un vertice fa parte di qualsiasi componente fortemente connesso var isScc = new List (n+1);  for (int i = 0; i< n + 1; i++)  {  isScc.Add(0);  }  var adj = new List>(n+1);  for (int i = 0; i< n + 1; i++)  {  adj.Add(new List ());  } 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 ();  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> bordi = nuova lista> { nuova lista { 1, 3 }, nuova lista { 1, 4 }, nuova lista { 2, 1 }, nuova lista { 3, 2 }, nuova lista { 4, 5 } };  Elenco> ans = obj.FindSCC(V, bordi);  Console.WriteLine('I componenti fortemente connessi sono:');  foreach (var x in ans) { foreach (var y in x) { Console.Write(y + ' ');  } Console.WriteLine();  } } } // Questo codice è stato fornito da shivamgupta310570>

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(() =>[]);  per (sia 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>