Gegeven een Weighted Directed Acyclic Graph (DAG) en een bronhoekpunt daarin, vind je de langste afstanden van het bronpunt tot alle andere hoekpunten in de gegeven grafiek.
We hebben al besproken hoe we het kunnen vinden Langste pad in gerichte acyclische grafiek (DAG) in Set 1. In dit bericht zullen we een andere interessante oplossing bespreken om het langste pad van DAG te vinden dat een algoritme gebruikt voor het vinden Kortste pad in een DAG .
Het idee is om negeer de gewichten van het pad en vind het kortste pad in de grafiek . Een langste pad tussen twee gegeven hoekpunten s en t in een gewogen grafiek G is hetzelfde als een kortste pad in een grafiek G' afgeleid van G door elk gewicht in zijn negatie te veranderen. Dus als de kortste paden in G' gevonden kunnen worden, dan kunnen de langste paden ook in G gevonden worden.
Hieronder vindt u het stapsgewijze proces voor het vinden van de langste paden -
java tostring
We veranderen het gewicht van elke rand van een bepaalde grafiek in de negatie ervan en initialiseren de afstanden tot alle hoekpunten als oneindig en de afstand tot de bron als 0. Vervolgens vinden we een topologische sortering van de grafiek die een lineaire ordening van de grafiek vertegenwoordigt. Wanneer we een hoekpunt u in topologische volgorde beschouwen, is het gegarandeerd dat we elke binnenkomende rand ervan hebben overwogen. dat wil zeggen dat we het kortste pad naar dat hoekpunt al hebben gevonden en we kunnen die informatie gebruiken om het kortere pad van alle aangrenzende hoekpunten bij te werken. Zodra we de topologische volgorde hebben, verwerken we één voor één alle hoekpunten in topologische volgorde. Voor elk hoekpunt dat wordt verwerkt, werken we de afstanden van het aangrenzende hoekpunt bij met behulp van de kortste afstand van het huidige hoekpunt tot het bronpunt en het randgewicht ervan. d.w.z.
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)
Zodra we alle kortste paden vanaf het bronpunt hebben gevonden, zijn de langste paden slechts een ontkenning van de kortste paden.
Hieronder vindt u de implementatie van de bovenstaande aanpak:
char naar Java met gehele getallenC++
// A C++ program to find single source longest distances // in a DAG #include using namespace std; // 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 { int v; int weight; public: AdjListNode(int _v int _w) { v = _v; weight = _w; } int getV() { return v; } int getWeight() { return weight; } }; // Graph class represents a directed graph using adjacency // list representation class Graph { int V; // No. of vertices // Pointer to an array containing adjacency lists list<AdjListNode>* adj; // This function uses DFS void longestPathUtil(int vector<bool> & stack<int> &); public: Graph(int); // Constructor ~Graph(); // Destructor // function to add an edge to graph void addEdge(int int int); void longestPath(int); }; Graph::Graph(int V) // Constructor { this->V = V; adj = new list<AdjListNode>[V]; } Graph::~Graph() // Destructor { delete[] adj; } void Graph::addEdge(int u int v int weight) { AdjListNode node(v weight); adj[u].push_back(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/ void Graph::longestPathUtil(int v vector<bool> &visited stack<int> &Stack) { // Mark the current node as visited visited[v] = true; // Recur for all the vertices adjacent to this vertex for (AdjListNode node : 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 void Graph::longestPath(int s) { // Initialize distances to all vertices as infinite and // distance to source as 0 int dist[V]; for (int i = 0; i < V; i++) dist[i] = INT_MAX; dist[s] = 0; stack<int> Stack; // Mark all the vertices as not visited vector<bool> visited(V false); for (int i = 0; i < V; i++) if (visited[i] == false) longestPathUtil(i visited Stack); // Process vertices in topological order while (!Stack.empty()) { // Get the next vertex from topological order int u = Stack.top(); Stack.pop(); if (dist[u] != INT_MAX) { // Update distances of all adjacent vertices // (edge from u -> v exists) for (AdjListNode v : 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] == INT_MAX) cout << 'INT_MIN '; else cout << (dist[i] * -1) << ' '; } } // Driver code int main() { Graph g(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; cout << 'Following are longest distances from ' << 'source vertex ' << s << ' n'; g.longestPath(s); return 0; }
Python3 # 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# // 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.
Java // 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
JavaScript 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
Uitvoer
Following are longest distances from source vertex 1 INT_MIN 0 2 9 8 10
Tijdcomplexiteit : Tijdscomplexiteit van topologische sortering is O(V+E). Nadat de topologische volgorde is gevonden, verwerkt het algoritme alle hoekpunten en voert het voor elk hoekpunt een lus uit voor alle aangrenzende hoekpunten. Omdat het totale aantal aangrenzende hoekpunten in een grafiek O(E) is, loopt de binnenste lus O(V + E) keer. Daarom is de totale tijdscomplexiteit van dit algoritme O(V + E).
Ruimtecomplexiteit:
De ruimtecomplexiteit van het bovenstaande algoritme is O(V). We slaan de uitvoerarray en een stapel op voor topologische sortering.