Camí més llarg d'un gràfic acíclic dirigit | Set 2
Donat un gràfic acíclic dirigit ponderat (DAG) i un vèrtex font, trobeu les distàncies més llargues des del vèrtex font fins a tots els altres vèrtexs del gràfic donat.
Ja hem comentat com podem trobar Camí més llarg del gràfic acíclic dirigit (DAG) al conjunt 1. En aquesta publicació parlarem d'una altra solució interessant per trobar el camí més llarg de DAG que utilitza un algorisme per trobar El camí més curt d'un DAG .
La idea és fer nega els pesos del camí i troba el camí més curt al gràfic . Un camí més llarg entre dos vèrtexs donats s i t en un gràfic ponderat G és el mateix que un camí més curt en un gràfic G' derivat de G canviant cada pes a la seva negació. Per tant, si els camins més curts es poden trobar a G', els camins més llargs també es poden trobar a G.
A continuació es mostra el procés pas a pas per trobar els camins més llargs:
Canviem el pes de cada aresta d'un gràfic donat a la seva negació i inicialitzem les distàncies a tots els vèrtexs com a infinites i la distància a la font com a 0, després trobem una ordenació topològica del gràfic que representa un ordre lineal del gràfic. Quan considerem un vèrtex u en ordre topològic, es garanteix que hem considerat totes les arestes entrants. és a dir, ja hem trobat el camí més curt cap a aquest vèrtex i podem utilitzar aquesta informació per actualitzar el camí més curt de tots els seus vèrtexs adjacents. Un cop tenim l'ordre topològic, processem un per un tots els vèrtexs en ordre topològic. Per a cada vèrtex que es processa, actualitzem les distàncies del seu vèrtex adjacent utilitzant la distància més curta del vèrtex actual des del vèrtex d'origen i el seu pes de vora. és a dir
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)
Un cop hem trobat tots els camins més curts des del vèrtex font, els camins més llargs seran només la negació dels camins més curts.
A continuació es mostra la implementació de l'enfocament anterior:
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 adj Stack 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 adj V 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
Sortida
Following are longest distances from source vertex 1 INT_MIN 0 2 9 8 10
Complexitat temporal : La complexitat temporal de l'ordenació topològica és O (V + E). Després de trobar l'ordre topològic, l'algorisme processa tots els vèrtexs i per a cada vèrtex executa un bucle per a tots els vèrtexs adjacents. Com que el total de vèrtexs adjacents en un gràfic és O(E), el bucle interior s'executa O(V + E) vegades. Per tant, la complexitat temporal global d'aquest algorisme és O(V + E).
Complexitat espacial:
La complexitat espacial de l'algorisme anterior és O (V). Estem emmagatzemant la matriu de sortida i una pila per a l'ordenació topològica.
