Givet en vægtet dirigeret acyklisk graf (DAG) og et kildepunkt i den, find de længste afstande fra kilden til alle andre hjørner i den givne graf.
Vi har allerede diskuteret, hvordan vi kan finde Længste sti i Directed Acyclic Graph (DAG) i sæt 1. I dette indlæg vil vi diskutere en anden interessant løsning til at finde den længste vej til DAG, der bruger algoritme til at finde Korteste vej i en DAG .
Tanken er at ophæve vægten af stien og find den korteste vej i grafen . En længste vej mellem to givne hjørner s og t i en vægtet graf G er det samme som en korteste vej i en graf G' afledt af G ved at ændre hver vægt til dens negation. Derfor, hvis de korteste veje kan findes i G', kan de længste stier også findes i G.
Nedenfor er trin for trin processen med at finde de længste stier -
Vi ændrer vægten af hver kant af en given graf til dens negation og initialiserer afstande til alle toppunkter som uendelige og afstand til kilde som 0, så finder vi en topologisk sortering af grafen, som repræsenterer en lineær rækkefølge af grafen. Når vi betragter et toppunkt u i topologisk rækkefølge, er det garanteret, at vi har overvejet hver indkommende kant til den. dvs. vi har allerede fundet den korteste vej til det toppunkt, og vi kan bruge den information til at opdatere den kortere vej for alle dens tilstødende toppunkter. Når vi først har topologisk orden, behandler vi en efter en alle toppunkter i topologisk rækkefølge. For hvert toppunkt, der behandles, opdaterer vi afstande af dets tilstødende toppunkt ved hjælp af den korteste afstand af det aktuelle toppunkt fra kildens toppunkt og dets kantvægt. dvs.
for every adjacent vertex v of every vertex u in topological order if (dist[v] > dist[u] + weight(u v)) dist[v] = dist[u] + weight(u v)
Når vi har fundet alle de korteste veje fra kildens toppunkt, vil de længste veje kun være negation af korteste veje.
Nedenfor er implementeringen af ovenstående tilgang:
C++// A C++ program to find single source longest distances // in a DAG #include
# A Python3 program to find single source # longest distances in a DAG import sys def addEdge(u v w): global adj adj[u].append([v w]) # A recursive function used by longestPath. # See below link for details. # https:#www.geeksforgeeks.org/topological-sorting/ def longestPathUtil(v): global visited adjStack visited[v] = 1 # Recur for all the vertices adjacent # to this vertex for node in adj[v]: if (not visited[node[0]]): longestPathUtil(node[0]) # Push current vertex to stack which # stores topological sort Stack.append(v) # The function do Topological Sort and finds # longest distances from given source vertex def longestPath(s): # Initialize distances to all vertices # as infinite and global visited Stack adjV dist = [sys.maxsize for i in range(V)] # for (i = 0 i < V i++) # dist[i] = INT_MAX dist[s] = 0 for i in range(V): if (visited[i] == 0): longestPathUtil(i) # print(Stack) while (len(Stack) > 0): # Get the next vertex from topological order u = Stack[-1] del Stack[-1] if (dist[u] != sys.maxsize): # Update distances of all adjacent vertices # (edge from u -> v exists) for v in adj[u]: # Consider negative weight of edges and # find shortest path if (dist[v[0]] > dist[u] + v[1] * -1): dist[v[0]] = dist[u] + v[1] * -1 # Print the calculated longest distances for i in range(V): if (dist[i] == sys.maxsize): print('INT_MIN ' end = ' ') else: print(dist[i] * (-1) end = ' ') # Driver code if __name__ == '__main__': V = 6 visited = [0 for i in range(7)] Stack = [] adj = [[] for i in range(7)] addEdge(0 1 5) addEdge(0 2 3) addEdge(1 3 6) addEdge(1 2 2) addEdge(2 4 4) addEdge(2 5 2) addEdge(2 3 7) addEdge(3 5 1) addEdge(3 4 -1) addEdge(4 5 -2) s = 1 print('Following are longest distances from source vertex' s) longestPath(s) # This code is contributed by mohit kumar 29
// C# program to find single source longest distances // in a DAG using System; using System.Collections.Generic; // Graph is represented using adjacency list. Every node of // adjacency list contains vertex number of the vertex to // which edge connects. It also contains weight of the edge class AdjListNode { private int v; private int weight; public AdjListNode(int _v int _w) { v = _v; weight = _w; } public int getV() { return v; } public int getWeight() { return weight; } } // Graph class represents a directed graph using adjacency // list representation class Graph { private int V; // No. of vertices // Pointer to an array containing adjacency lists private List<AdjListNode>[] adj; public Graph(int v) // Constructor { V = v; adj = new List<AdjListNode>[ v ]; for (int i = 0; i < v; i++) adj[i] = new List<AdjListNode>(); } public void AddEdge(int u int v int weight) { AdjListNode node = new AdjListNode(v weight); adj[u].Add(node); // Add v to u's list } // A recursive function used by longestPath. See below // link for details. // https://www.geeksforgeeks.org/dsa/topological-sorting/ private void LongestPathUtil(int v bool[] visited Stack<int> stack) { // Mark the current node as visited visited[v] = true; // Recur for all the vertices adjacent to this // vertex foreach(AdjListNode node in adj[v]) { if (!visited[node.getV()]) LongestPathUtil(node.getV() visited stack); } // Push current vertex to stack which stores // topological sort stack.Push(v); } // The function do Topological Sort and finds longest // distances from given source vertex public void LongestPath(int s) { // Initialize distances to all vertices as infinite // and distance to source as 0 int[] dist = new int[V]; for (int i = 0; i < V; i++) dist[i] = Int32.MaxValue; dist[s] = 0; Stack<int> stack = new Stack<int>(); // Mark all the vertices as not visited bool[] visited = new bool[V]; for (int i = 0; i < V; i++) { if (visited[i] == false) LongestPathUtil(i visited stack); } // Process vertices in topological order while (stack.Count > 0) { // Get the next vertex from topological order int u = stack.Pop(); if (dist[u] != Int32.MaxValue) { // Update distances of all adjacent vertices // (edge from u -> v exists) foreach(AdjListNode v in adj[u]) { // consider negative weight of edges and // find shortest path if (dist[v.getV()] > dist[u] + v.getWeight() * -1) dist[v.getV()] = dist[u] + v.getWeight() * -1; } } } // Print the calculated longest distances for (int i = 0; i < V; i++) { if (dist[i] == Int32.MaxValue) Console.Write('INT_MIN '); else Console.Write('{0} ' dist[i] * -1); } Console.WriteLine(); } } public class GFG { // Driver code static void Main(string[] args) { Graph g = new Graph(6); g.AddEdge(0 1 5); g.AddEdge(0 2 3); g.AddEdge(1 3 6); g.AddEdge(1 2 2); g.AddEdge(2 4 4); g.AddEdge(2 5 2); g.AddEdge(2 3 7); g.AddEdge(3 5 1); g.AddEdge(3 4 -1); g.AddEdge(4 5 -2); int s = 1; Console.WriteLine( 'Following are longest distances from source vertex {0} ' s); g.LongestPath(s); } } // This code is contributed by cavi4762.
// A Java program to find single source longest distances // in a DAG import java.util.*; // Graph is represented using adjacency list. Every // node of adjacency list contains vertex number of // the vertex to which edge connects. It also // contains weight of the edge class AdjListNode { private int v; private int weight; AdjListNode(int _v int _w) { v = _v; weight = _w; } int getV() { return v; } int getWeight() { return weight; } } // Class to represent a graph using adjacency list // representation public class GFG { int V; // No. of vertices' // Pointer to an array containing adjacency lists ArrayList<AdjListNode>[] adj; @SuppressWarnings('unchecked') GFG(int V) // Constructor { this.V = V; adj = new ArrayList[V]; for (int i = 0; i < V; i++) { adj[i] = new ArrayList<>(); } } void addEdge(int u int v int weight) { AdjListNode node = new AdjListNode(v weight); adj[u].add(node); // Add v to u's list } // A recursive function used by longestPath. See // below link for details https:// // www.geeksforgeeks.org/topological-sorting/ void topologicalSortUtil(int v boolean visited[] Stack<Integer> stack) { // Mark the current node as visited visited[v] = true; // Recur for all the vertices adjacent to this // vertex for (int i = 0; i < adj[v].size(); i++) { AdjListNode node = adj[v].get(i); if (!visited[node.getV()]) topologicalSortUtil(node.getV() visited stack); } // Push current vertex to stack which stores // topological sort stack.push(v); } // The function to find Smallest distances from a // given vertex. It uses recursive // topologicalSortUtil() to get topological sorting. void longestPath(int s) { Stack<Integer> stack = new Stack<Integer>(); int dist[] = new int[V]; // Mark all the vertices as not visited boolean visited[] = new boolean[V]; for (int i = 0; i < V; i++) visited[i] = false; // Call the recursive helper function to store // Topological Sort starting from all vertices // one by one for (int i = 0; i < V; i++) if (visited[i] == false) topologicalSortUtil(i visited stack); // Initialize distances to all vertices as // infinite and distance to source as 0 for (int i = 0; i < V; i++) dist[i] = Integer.MAX_VALUE; dist[s] = 0; // Process vertices in topological order while (stack.isEmpty() == false) { // Get the next vertex from topological // order int u = stack.peek(); stack.pop(); // Update distances of all adjacent vertices if (dist[u] != Integer.MAX_VALUE) { for (AdjListNode v : adj[u]) { if (dist[v.getV()] > dist[u] + v.getWeight() * -1) dist[v.getV()] = dist[u] + v.getWeight() * -1; } } } // Print the calculated longest distances for (int i = 0; i < V; i++) if (dist[i] == Integer.MAX_VALUE) System.out.print('INF '); else System.out.print(dist[i] * -1 + ' '); } // Driver program to test above functions public static void main(String args[]) { // Create a graph given in the above diagram. // Here vertex numbers are 0 1 2 3 4 5 with // following mappings: // 0=r 1=s 2=t 3=x 4=y 5=z GFG g = new GFG(6); g.addEdge(0 1 5); g.addEdge(0 2 3); g.addEdge(1 3 6); g.addEdge(1 2 2); g.addEdge(2 4 4); g.addEdge(2 5 2); g.addEdge(2 3 7); g.addEdge(3 5 1); g.addEdge(3 4 -1); g.addEdge(4 5 -2); int s = 1; System.out.print( 'Following are longest distances from source vertex ' + s + ' n'); g.longestPath(s); } } // This code is contributed by Prithi_Dey
class AdjListNode { constructor(v weight) { this.v = v; this.weight = weight; } getV() { return this.v; } getWeight() { return this.weight; } } class GFG { constructor(V) { this.V = V; this.adj = new Array(V); for (let i = 0; i < V; i++) { this.adj[i] = new Array(); } } addEdge(u v weight) { let node = new AdjListNode(v weight); this.adj[u].push(node); } topologicalSortUtil(v visited stack) { visited[v] = true; for (let i = 0; i < this.adj[v].length; i++) { let node = this.adj[v][i]; if (!visited[node.getV()]) { this.topologicalSortUtil(node.getV() visited stack); } } stack.push(v); } longestPath(s) { let stack = new Array(); let dist = new Array(this.V); let visited = new Array(this.V); for (let i = 0; i < this.V; i++) { visited[i] = false; } for (let i = 0; i < this.V; i++) { if (!visited[i]) { this.topologicalSortUtil(i visited stack); } } for (let i = 0; i < this.V; i++) { dist[i] = Number.MAX_SAFE_INTEGER; } dist[s] = 0; let u = stack.pop(); while (stack.length > 0) { u = stack.pop(); if (dist[u] !== Number.MAX_SAFE_INTEGER) { for (let v of this.adj[u]) { if (dist[v.getV()] > dist[u] + v.getWeight() * -1) { dist[v.getV()] = dist[u] + v.getWeight() * -1; } } } } for (let i = 0; i < this.V; i++) { if (dist[i] === Number.MAX_SAFE_INTEGER) { console.log('INF'); } else { console.log(dist[i] * -1); } } } } let g = new GFG(6); g.addEdge(0 1 5); g.addEdge(0 2 3); g.addEdge(1 3 6); g.addEdge(1 2 2); g.addEdge(2 4 4); g.addEdge(2 5 2); g.addEdge(2 3 7); g.addEdge(3 5 1); g.addEdge(3 4 -1); g.addEdge(4 5 -2); console.log('Longest distances from the vertex 1 : '); g.longestPath(1); //this code is contributed by devendra
Produktion
Following are longest distances from source vertex 1 INT_MIN 0 2 9 8 10
Tidskompleksitet : Tidskompleksiteten af topologisk sortering er O(V + E). Efter at have fundet topologisk rækkefølge behandler algoritmen alle toppunkter og for hvert toppunkt kører den en løkke for alle tilstødende toppunkter. Da de samlede tilstødende hjørner i en graf er O(E), løber den indre sløjfe O(V + E) gange. Derfor er den samlede tidskompleksitet af denne algoritme O(V + E).
Rumkompleksitet:
Rumkompleksiteten af ovenstående algoritme er O(V). Vi gemmer output-arrayet og en stak til topologisk sortering.