S obzirom na usmjereni Eulerov graf zadatak je ispisati Eulerov krug . Eulerov krug je staza koja prolazi svaki rub grafa točno jednom i staza završava na početnom vrhu.
Bilješka: Zadani graf sadrži Eulerov krug.
Primjer:
Ulaz: usmjereni graf
Izlaz: 0 3 4 0 2 1 0
Preduvjeti:
- Razgovarali smo o problem otkrivanja je li dati graf Eulerov ili nije za neusmjereni graf
- Uvjeti za Eulerov krug u usmjerenom grpagu: (1) Svi vrhovi pripadaju jednoj jako povezanoj komponenti. (2) Svi vrhovi imaju isti ulazni i izlazni stupanj. Imajte na umu da je za neusmjereni graf uvjet drugačiji (svi vrhovi imaju paran stupanj)
Pristup:
- Odaberite bilo koji početni vrh v i slijedite trag bridova od tog vrha do povratka u v. Nije moguće zapeti ni na jednom vrhu osim v jer indegree i outdegree svakog vrha moraju biti isti kada trag uđe u drugi vrh w mora postojati neiskorišten brid koji izlazi iz w. Ovako formirana tura je zatvorena tura ali možda neće pokriti sve vrhove i bridove početnog grafa.
- Sve dok postoji vrh u koji pripada trenutnom obilasku, ali ima susjedne bridove koji nisu dio obilaska, započnite drugi trag od u prateći neiskorištene bridove do povratka na u i pridružite tako formirani obilazak prethodnom obilasku.
Ilustracija:
Uzimamo primjer gornjeg grafikona s 5 čvorova: adj = {{2 3} {0} {1} {4} {0}}.
- Počnite od vrha 0 :
- Trenutni put: [0]
- Krug: []
- Vrh 0 → 3 :
- Trenutni put: [0 3]
- Krug: []
- Verteks 3 → 4 :
- Trenutni put: [0 3 4]
- Krug: []
- Vrh 4 → 0 :
- Trenutni put: [0 3 4 0]
- Krug: []
- Vrh 0 → 2 :
- Trenutni put: [0 3 4 0 2]
- Krug: []
- Vrh 2 → 1 :
- Trenutni put: [0 3 4 0 2 1]
- Krug: []
- Vrh 1 → 0 :
- Trenutni put: [0 3 4 0 2 1 0]
- Krug: []
- Povratak do vrha 0 : Dodajte 0 krugu.
- Trenutni put: [0 3 4 0 2 1]
- Krug: [0]
- Povratak do vrha 1 : Dodajte 1 krugu.
- Trenutni put: [0 3 4 0 2]
- Krug: [0 1]
- Povratak do vrha 2 : Dodajte 2 krugu.
- Trenutni put: [0 3 4 0]
- Krug: [0 1 2]
- Povratak do vrha 0 : Dodajte 0 krugu.
- Trenutni put: [0 3 4]
- Krug: [0 1 2 0]
- Povratak do vrha 4 : Dodajte 4 krugu.
- Trenutni put: [0 3]
- Krug: [0 1 2 0 4]
- Povratak do vrha 3 : Dodajte 3 krugu.
- Trenutni put: [0]
- Krug: [0 1 2 0 4 3]
- Povratak do vrha 0 : Dodajte 0 krugu.
- Trenutačni put: []
- Krug: [0 1 2 0 4 3 0]
U nastavku je implementacija gornjeg pristupa:
C++// C++ program to print Eulerian circuit in given // directed graph using Hierholzer algorithm #include
// Java program to print Eulerian circuit in given // directed graph using Hierholzer algorithm import java.util.*; class GfG { // Function to print Eulerian circuit static List<Integer> printCircuit(List<List<Integer>> adj) { int n = adj.size(); if (n == 0) return new ArrayList<>(); // Maintain a stack to keep vertices // We can start from any vertex here we start with 0 List<Integer> currPath = new ArrayList<>(); currPath.add(0); // list to store final circuit List<Integer> circuit = new ArrayList<>(); while (currPath.size() > 0) { int currNode = currPath.get(currPath.size() - 1); // If there's remaining edge in adjacency list // of the current vertex if (adj.get(currNode).size() > 0) { // Find and remove the next vertex that is // adjacent to the current vertex int nextNode = adj.get(currNode).get(adj.get(currNode).size() - 1); adj.get(currNode).remove(adj.get(currNode).size() - 1); // Push the new vertex to the stack currPath.add(nextNode); } // back-track to find remaining circuit else { // Remove the current vertex and // put it in the circuit circuit.add(currPath.get(currPath.size() - 1)); currPath.remove(currPath.size() - 1); } } // reverse the result vector Collections.reverse(circuit); return circuit; } public static void main(String[] args) { List<List<Integer>> adj = new ArrayList<>(); adj.add(new ArrayList<>(Arrays.asList(2 3))); adj.add(new ArrayList<>(Arrays.asList(0))); adj.add(new ArrayList<>(Arrays.asList(1))); adj.add(new ArrayList<>(Arrays.asList(4))); adj.add(new ArrayList<>(Arrays.asList(0))); List<Integer> ans = printCircuit(adj); for (int v : ans) System.out.print(v + ' '); System.out.println(); } }
# Python program to print Eulerian circuit in given # directed graph using Hierholzer algorithm # Function to print Eulerian circuit def printCircuit(adj): n = len(adj) if n == 0: return [] # Maintain a stack to keep vertices # We can start from any vertex here we start with 0 currPath = [0] # list to store final circuit circuit = [] while len(currPath) > 0: currNode = currPath[-1] # If there's remaining edge in adjacency list # of the current vertex if len(adj[currNode]) > 0: # Find and remove the next vertex that is # adjacent to the current vertex nextNode = adj[currNode].pop() # Push the new vertex to the stack currPath.append(nextNode) # back-track to find remaining circuit else: # Remove the current vertex and # put it in the circuit circuit.append(currPath.pop()) # reverse the result vector circuit.reverse() return circuit if __name__ == '__main__': adj = [[2 3] [0] [1] [4] [0]] ans = printCircuit(adj) for v in ans: print(v end=' ') print()
// C# program to print Eulerian circuit in given // directed graph using Hierholzer algorithm using System; using System.Collections.Generic; class GfG { // Function to print Eulerian circuit static List<int> printCircuit(List<List<int>> adj) { int n = adj.Count; if (n == 0) return new List<int>(); // Maintain a stack to keep vertices // We can start from any vertex here we start with 0 List<int> currPath = new List<int> { 0 }; // list to store final circuit List<int> circuit = new List<int>(); while (currPath.Count > 0) { int currNode = currPath[currPath.Count - 1]; // If there's remaining edge in adjacency list // of the current vertex if (adj[currNode].Count > 0) { // Find and remove the next vertex that is // adjacent to the current vertex int nextNode = adj[currNode][adj[currNode].Count - 1]; adj[currNode].RemoveAt(adj[currNode].Count - 1); // Push the new vertex to the stack currPath.Add(nextNode); } // back-track to find remaining circuit else { // Remove the current vertex and // put it in the circuit circuit.Add(currPath[currPath.Count - 1]); currPath.RemoveAt(currPath.Count - 1); } } // reverse the result vector circuit.Reverse(); return circuit; } static void Main(string[] args) { List<List<int>> adj = new List<List<int>> { new List<int> { 2 3 } new List<int> { 0 } new List<int> { 1 } new List<int> { 4 } new List<int> { 0 } }; List<int> ans = printCircuit(adj); foreach (int v in ans) { Console.Write(v + ' '); } Console.WriteLine(); } }
// JavaScript program to print Eulerian circuit in given // directed graph using Hierholzer algorithm // Function to print Eulerian circuit function printCircuit(adj) { let n = adj.length; if (n === 0) return []; // Maintain a stack to keep vertices // We can start from any vertex here we start with 0 let currPath = [0]; // list to store final circuit let circuit = []; while (currPath.length > 0) { let currNode = currPath[currPath.length - 1]; // If there's remaining edge in adjacency list // of the current vertex if (adj[currNode].length > 0) { // Find and remove the next vertex that is // adjacent to the current vertex let nextNode = adj[currNode].pop(); // Push the new vertex to the stack currPath.push(nextNode); } // back-track to find remaining circuit else { // Remove the current vertex and // put it in the circuit circuit.push(currPath.pop()); } } // reverse the result vector circuit.reverse(); return circuit; } let adj = [[2 3] [0] [1] [4] [0]]; let ans = printCircuit(adj); for (let v of ans) { console.log(v ' '); }
Izlaz
0 3 4 0 2 1 0
Vremenska složenost: O(V + E) gdje je V broj vrhova, a E broj bridova u grafu. Razlog za to je taj što algoritam izvodi dubinsko pretraživanje (DFS) i posjećuje svaki vrh i svaki rub točno jednom. Dakle, za svaki vrh je potrebno O(1) vremena da ga se obiđe i za svaki brid je potrebno O(1) vremena da ga se obiđe.
Složenost prostora : O(V + E) kao algoritam koristi stog za pohranjivanje trenutne putanje i popis za pohranjivanje konačnog kruga. Maksimalna veličina hrpe može biti V + E u najgorem slučaju, tako da je složenost prostora O(V + E).
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