एक ग्राफ एक पेड़ है यदि इसमें कोई चक्र नहीं है। यह एक C++ प्रोग्राम है जो यह जांचने के लिए है कि निर्देशित ग्राफ ट्री है या नहीं DFS का उपयोग कर रहा है।
एल्गोरिदम
Begin function cyclicUtil() : a) Mark the current node as visited and part of recursion stack b) Recur for all the vertices adjacent to this vertex. c) Remove the vertex from recursion stack. function cyclic() : a) Mark all the vertices as not visited and not part of recursion stack b) Call the CyclicUtill() function to detect cycle in different trees End
उदाहरण
#include<iostream> #include <list> #include <limits.h> using namespace std; class G { int n; list<int> *adj; //contain adjacent list. bool CyclicUtil(int v, bool visited[], bool *rs); public: G(int V); // Constructor void addEd(int v, int w); bool cyclic(); }; G::G(int n) { this->n = n; adj = new list<int> [n]; } void G::addEd(int v, int u) //to add edges in the graph { adj[v].push_back(u); //add u to v’s list } bool G::CyclicUtil(int v, bool visited[], bool *recurS) { if (visited[v] == false) { visited[v] = true; //Mark the current node as visited and part of recursion stack recurS[v] = true; // 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] && CyclicUtil(*i, visited, recurS)) return true; else if (recurS[*i]) return true; } } recurS[v] = false; //Remove the vertex from recursion stack. return false; } //check if the graph is tree or not bool G::cyclic() { //Mark all the vertices as not visited and not part of recursion stack bool *visited = new bool[n]; bool *recurS = new bool[n]; for (int i = 0; i < n; i++) { visited[i] = false; recurS[i] = false; } // Call the CyclicUtill() function to detect cycle in different trees for (int i = 0; i < n; i++) if (CyclicUtil(i, visited, recurS)) return true; return false; } int main() { G g(4); g.addEd(0, 2); g.addEd(1, 2); g.addEd(2, 0); g.addEd(3, 2); if (g.cyclic()) cout << "Directed Graph isn't a tree"; else cout << "Directed Graph is a tree"; return 0; }
आउटपुट
Directed Graph isn't a tree