Lexicographically Smallest Topological Ordering
Given a directed graph with N vertices and M edges that may contain cycles, the task is to find the lexicographically smallest topological ordering of the graph if it exists otherwise print -1 (if the graph has cycles).
Lexigraphically smallest topological ordering means that if two vertices in a graph do not have any incoming edge then the vertex with the smaller number should appear first in the ordering.
For Example, in the image below many topological orderings are possible e.g 5 2 3 4 0 1, 5 0 2 4 3 1.
But the smallest ordering is 4 5 0 2 3 1.
Examples:
Input:
Output: 4 5 0 2 3 1
Even though 5 4 0 2 3 1 is also a valid topological
ordering of the given graph but it is not
lexicographically smallest.
Approach: We will use Kahn’s algorithm for Topological Sorting with a modification. Instead of using a queue we will use a multiset to store the vertices to make sure that every time we pick a vertex it is the smallest possible of all. The overall Time complexity changes to 
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach#include<bits/stdc++.h>using namespace std;vector<vector<int>> adj;// function to add edge to the graphvoid addEdge(int x,int y){ adj[x].push_back(y);}// Function to print the required topological// sort of the given graphvoid topologicalSort(){ int V = adj.size(); // Create a vector to store indegrees of all // the vertices // Initialize all indegrees to 0 vector<int> in_degree(V, 0); // Traverse adjacency lists to fill indegrees of // vertices // This step takes O(V+E) time for (int u = 0; u < V; u++) { for (auto x: adj[u]) in_degree[x]++; } // Create a set and inserting all vertices with // indegree 0 multiset<int> s; for (int i = 0; i < V; i++) if (in_degree[i] == 0) s.insert(i); // Initialize count of visited vertices int cnt = 0; // Create a vector to store result (A topological // ordering of the vertices) vector<int> top_order; // One by one erase vertices from setand insert // adjacents if indegree of adjacent becomes 0 while (!s.empty()) { // Extract vertex with minimum number from multiset // and add it to topological order int u = *s.begin(); s.erase(s.begin()); top_order.push_back(u); // Iterate through all its neighbouring nodes // of erased node u and decrease their in-degree // by 1 for (auto x:adj[u]) // If in-degree becomes zero, add it to queue if (--in_degree[x] == 0) s.insert(x); cnt++; } // Check if there was a cycle if (cnt != V) { cout << -1; return; } // Print topological order for (int i = 0; i < top_order.size(); i++) cout << top_order[i] << " ";}int main(){ // number of vertices int v = 6; // adjacency matrix adj= vector<vector<int>>(v); addEdge(5,2); addEdge(5,0); addEdge(4,0); addEdge(4,1); addEdge(2,3); addEdge(3,1); // find required topological order topologicalSort();} |
4 5 0 2 3 1
Time Complexity: O(N)
Auxiliary Space: O(N)
