Glede na uteženi usmerjeni aciklični graf (DAG) in izvorno vozlišče v njem poiščite najdaljše razdalje od izvornega vozlišča do vseh drugih vozlišč v danem grafu.
Razpravljali smo že o tem, kako najti Najdaljša pot v usmerjenem acikličnem grafu (DAG) v nizu 1. V tej objavi bomo obravnavali še eno zanimivo rešitev za iskanje najdaljše poti DAG, ki uporablja algoritem za iskanje Najkrajša pot v DAG .
string.format java niz
Ideja je, da negirajte uteži poti in poiščite najkrajšo pot v grafu . Najdaljša pot med dvema danima vozliščema s in t v uteženem grafu G je enaka kot najkrajša pot v grafu G', ki je izpeljan iz G s spremembo vsake uteži na njeno negacijo. Torej, če lahko najkrajše poti najdemo v G', potem lahko tudi najdaljše poti najdemo v G.
Spodaj je korak za korakom postopek iskanja najdaljših poti -
Težo vsakega roba danega grafa spremenimo v njegovo negacijo in inicializiramo razdalje do vseh vozlišč kot neskončne in razdaljo do izvora kot 0, nato pa najdemo topološko razvrščanje grafa, ki predstavlja linearno urejenost grafa. Ko obravnavamo vozlišče u v topološkem redu, je zagotovljeno, da smo upoštevali vse prihajajoče robove do njega. tj. že smo našli najkrajšo pot do te točke in lahko uporabimo to informacijo za posodobitev krajše poti vseh njenih sosednjih točk. Ko imamo topološki red, enega za drugim obdelamo vsa vozlišča v topološkem redu. Za vsako oglišče, ki se obdeluje, posodobimo razdalje njegovega sosednjega oglišča z uporabo najkrajše razdalje trenutnega oglišča od izvornega oglišča in njegove robne teže. tj.
matriko v jeziku c
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)
Ko najdemo vse najkrajše poti iz izvornega vozlišča, bodo najdaljše poti samo negacija najkrajših poti.
Spodaj je izvedba zgornjega pristopa:
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
Izhod
Following are longest distances from source vertex 1 INT_MIN 0 2 9 8 10
Časovna zapletenost : Časovna zahtevnost topološkega razvrščanja je O(V + E). Po najdbi topološkega reda algoritem obdela vsa oglišča in za vsako oglišče izvede zanko za vsa sosednja oglišča. Ker je skupno število sosednjih vozlišč v grafu O(E), se notranja zanka izvede O(V + E)-krat. Zato je skupna časovna kompleksnost tega algoritma O(V + E).
c++ razcepni niz
Kompleksnost prostora:
Prostorska kompleksnost zgornjega algoritma je O(V). Shranjujemo izhodno polje in sklad za topološko razvrščanje.