Atsižvelgiant į svertinį nukreiptą aciklinį grafiką (DAG) ir jame esančią šaltinio viršūnę, suraskite didžiausius atstumus nuo šaltinio viršūnės iki visų kitų duotojo grafiko viršūnių.
char į eilutę
Jau aptarėme, kaip galime rasti Ilgiausias kelias nukreiptoje aciklinėje diagramoje (DAG) 1 rinkinyje. Šiame įraše aptarsime kitą įdomų sprendimą, kaip rasti ilgiausią DAG kelią, kuris naudoja algoritmą Trumpiausias kelias DAG .
Idėja yra paneigti kelio svorius ir rasti trumpiausią kelią grafike . Ilgiausias kelias tarp dviejų duotųjų viršūnių s ir t svertiniame grafe G yra tas pats, kas trumpiausias kelias grafe G', gaunamas iš G, pakeičiant kiekvieną svorį į jo neigimą. Todėl jei trumpiausius kelius galima rasti G, tai ilgiausius kelius taip pat galima rasti G.
Žemiau pateikiamas žingsnis po žingsnio ilgiausių kelių paieškos procesas -
Pakeičiame kiekvieno duoto grafo krašto svorį į jo neigimą ir inicijuojame atstumus iki visų viršūnių kaip begalinius ir atstumą iki šaltinio kaip 0, tada randame topologinį grafiko rūšiavimą, kuris parodo tiesinę grafiko tvarką. Kai vertiname viršūnę u topologine tvarka, garantuojame, kad atsižvelgėme į kiekvieną įeinančią jos briauną. y., mes jau radome trumpiausią kelią į tą viršūnę ir galime naudoti šią informaciją norėdami atnaujinti trumpesnį visų gretimų viršūnių kelią. Kai turime topologinę tvarką, po vieną apdorojame visas viršūnes topologine tvarka. Kiekvienai apdorojamai viršūnei mes atnaujiname gretimos viršūnės atstumus, naudodami trumpiausią dabartinės viršūnės atstumą nuo šaltinio viršūnės ir jos krašto svorį. t.y.
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)
Kai mes surasime visus trumpiausius kelius iš šaltinio viršūnės, ilgiausi keliai bus tik trumpiausių kelių neigimas.
Žemiau pateikiamas pirmiau minėto metodo įgyvendinimas:
rūšiuoti masyvo sąrašąC++
// 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
Išvestis
Following are longest distances from source vertex 1 INT_MIN 0 2 9 8 10
Laiko sudėtingumas : Topologinio rūšiavimo laiko sudėtingumas yra O (V + E). Suradęs topologinę tvarką, algoritmas apdoroja visas viršūnes ir kiekvienai viršūnei paleidžia kilpą visoms gretimoms viršūnėms. Kadangi visos gretimų viršūnių grafe yra O(E), vidinė kilpa vyksta O(V + E) kartus. Todėl bendras šio algoritmo laiko sudėtingumas yra O(V + E).
Erdvės sudėtingumas:
Aukščiau pateikto algoritmo erdvės sudėtingumas yra O(V). Saugome išvesties masyvą ir krūvą topologiniam rūšiavimui.