Étant donné un graphique acyclique dirigé pondéré (DAG) et un sommet source, trouvez les distances les plus longues entre le sommet source et tous les autres sommets du graphique donné.
Nous avons déjà discuté de la façon dont nous pouvons trouver Chemin le plus long dans un graphique acyclique dirigé (DAG) dans l'ensemble 1. Dans cet article, nous discuterons d'une autre solution intéressante pour trouver le chemin le plus long du DAG qui utilise un algorithme pour trouver Le chemin le plus court dans un DAG .
L'idée est de annuler les poids du chemin et trouver le chemin le plus court dans le graphique . Un chemin le plus long entre deux sommets donnés s et t dans un graphe pondéré G est la même chose qu'un chemin le plus court dans un graphe G' dérivé de G en changeant chaque poids en sa négation. Par conséquent, si les chemins les plus courts peuvent être trouvés dans G', alors les chemins les plus longs peuvent également être trouvés dans G.
Vous trouverez ci-dessous le processus étape par étape pour trouver les chemins les plus longs -
Nous changeons le poids de chaque arête d'un graphe donné en sa négation et initialisons les distances à tous les sommets comme infinies et la distance à la source comme 0, puis nous trouvons un tri topologique du graphe qui représente un ordre linéaire du graphe. Lorsque nous considérons un sommet u dans l’ordre topologique, il est garanti que nous avons pris en compte toutes les arêtes entrantes. c'est-à-dire que nous avons déjà trouvé le chemin le plus court vers ce sommet et nous pouvons utiliser ces informations pour mettre à jour le chemin le plus court de tous ses sommets adjacents. Une fois que nous avons l'ordre topologique, nous traitons un par un tous les sommets dans l'ordre topologique. Pour chaque sommet en cours de traitement, nous mettons à jour les distances de son sommet adjacent en utilisant la distance la plus courte du sommet actuel par rapport au sommet source et son poids de bord. c'est-à-dire
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)
Une fois que nous avons trouvé tous les chemins les plus courts à partir du sommet source, les chemins les plus longs ne seront que la négation des chemins les plus courts.
Vous trouverez ci-dessous la mise en œuvre de l’approche ci-dessus :
C++// 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
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Following are longest distances from source vertex 1 INT_MIN 0 2 9 8 10
Complexité temporelle : La complexité temporelle du tri topologique est O(V + E). Après avoir trouvé l'ordre topologique, l'algorithme traite tous les sommets et pour chaque sommet, il exécute une boucle pour tous les sommets adjacents. Comme le total des sommets adjacents dans un graphique est O(E), la boucle interne s'exécute O(V + E) fois. Par conséquent, la complexité temporelle globale de cet algorithme est O(V + E).
Complexité spatiale :
La complexité spatiale de l'algorithme ci-dessus est O(V). Nous stockons le tableau de sortie et une pile pour le tri topologique.