A DAG is given to us, we need to find maximum number of edges that can be added to this DAG, after which new graph still remain a DAG that means the reformed graph should have maximal number of edges, adding even single edge will create a cycle in graph.
The Output for above example should be following edges in any order. 4-2, 4-5, 4-3, 5-3, 5-1, 2-0, 2-1, 0-3, 0-1
As shown in above example, we have added all the edges in one direction only to save ourselves from making a cycle. This is the trick to solve this question. We sort all our nodes in topological order and create edges from node to all nodes to the right if not there already.
How can we say that, it is not possible to add any more edge? the reason is we have added all possible edges from left to right and if we want to add more edge we need to make that from right to left, but adding edge from right to left we surely create a cycle because its counter part left to right edge is already been added to graph and creating cycle is not what we needed.
So solution proceeds as follows, we consider the nodes in topological order and if any edge is not there from left to right, we will create it.
Below is the solution, we have printed all the edges that can be added to given DAG without making any cycle.
C++
// C++ program to find maximum edges after adding // which graph still remains a DAG#include <bits/stdc++.h>using namespace std;class Graph{ int V; // No. of vertices // Pointer to a list containing adjacency list list<int> *adj; // Vector to store indegree of vertices vector<int> indegree; // function returns a topological sort vector<int> topologicalSort();public: Graph(int V); // Constructor // function to add an edge to graph void addEdge(int v, int w); // Prints all edges that can be added without making any cycle void maximumEdgeAddtion();};// Constructor of graphGraph::Graph(int V){ this->V = V; adj = new list<int>[V]; // Initialising all indegree with 0 for (int i = 0; i < V; i++) indegree.push_back(0);}// Utility function to add edgevoid Graph::addEdge(int v, int w){ adj[v].push_back(w); // Add w to v's list. // increasing inner degree of w by 1 indegree[w]++;}// Main function to print maximum edges that can be addedvector<int> Graph::topologicalSort(){ vector<int> topological; queue<int> q; // In starting push all node with indegree 0 for (int i = 0; i < V; i++) if (indegree[i] == 0) q.push(i); while (!q.empty()) { int t = q.front(); q.pop(); // push the node into topological vector topological.push_back(t); // reducing indegree of adjacent vertices for (list<int>:: iterator j = adj[t].begin(); j != adj[t].end(); j++) { indegree[*j]--; // if indegree becomes 0, just push // into queue if (indegree[*j] == 0) q.push(*j); } } return topological;}// The function prints all edges that can be// added without making any cycle// It uses recursive topologicalSort()void Graph::maximumEdgeAddtion(){ bool *visited = new bool[V]; vector<int> topo = topologicalSort(); // looping for all nodes for (int i = 0; i < topo.size(); i++) { int t = topo[i]; // In below loop we mark the adjacent node of t for (list<int>::iterator j = adj[t].begin(); j != adj[t].end(); j++) visited[*j] = true; // In below loop unmarked nodes are printed for (int j = i + 1; j < topo.size(); j++) { // if not marked, then we can make an edge // between t and j if (!visited[topo[j]]) cout << t << "-" << topo[j] << " "; visited[topo[j]] = false; } }}// Driver code to test above methodsint main(){ // Create a graph given in the above diagram Graph g(6); g.addEdge(5, 2); g.addEdge(5, 0); g.addEdge(4, 0); g.addEdge(4, 1); g.addEdge(2, 3); g.addEdge(3, 1); g.maximumEdgeAddtion(); return 0;} |
Python3
# Python3 program to find maximum# edges after adding which graph # still remains a DAG class Graph: def __init__(self, V): # No. of vertices self.V = V # Pointer to a list containing # adjacency list self.adj = [[] for i in range(V)] # Vector to store indegree of vertices self.indegree = [0 for i in range(V)] # Utility function to add edge def addEdge(self, v, w): # Add w to v's list. self.adj[v].append(w) # Increasing inner degree of w by 1 self.indegree[w] += 1 # Main function to print maximum # edges that can be added def topologicalSort(self): topological = [] q = [] # In starting append all node # with indegree 0 for i in range(self.V): if (self.indegree[i] == 0): q.append(i) while (len(q) != 0): t = q[0] q.pop(0) # Append the node into topological # vector topological.append(t) # Reducing indegree of adjacent # vertices for j in self.adj[t]: self.indegree[j] -= 1 # If indegree becomes 0, just # append into queue if (self.indegree[j] == 0): q.append(j) return topological # The function prints all edges that # can be added without making any cycle # It uses recursive topologicalSort() def maximumEdgeAddtion(self): visited = [False for i in range(self.V)] topo = self.topologicalSort() # Looping for all nodes for i in range(len(topo)): t = topo[i] # In below loop we mark the # adjacent node of t for j in self.adj[t]: visited[j] = True # In below loop unmarked nodes # are printed for j in range(i + 1, len(topo)): # If not marked, then we can make # an edge between t and j if (not visited[topo[j]]): print(str(t) + '-' + str(topo[j]), end = ' ') visited[topo[j]] = False # Driver code if __name__=='__main__': # Create a graph given in the # above diagram g = Graph(6) g.addEdge(5, 2) g.addEdge(5, 0) g.addEdge(4, 0) g.addEdge(4, 1) g.addEdge(2, 3) g.addEdge(3, 1) g.maximumEdgeAddtion()# This code is contributed by rutvik_56 |
Output:
4-5, 4-2, 4-3, 5-3, 5-1, 2-0, 2-1, 0-3, 0-1
This article is contributed by Utkarsh Trivedi. 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
