Path Problems in Acyclic Graphs in Java Creation Code39 in Java Path Problems in Acyclic Graphs Path Problems in Acyclic GraphsPaint Code 3/9 In JavaUsing Barcode creator for Java Control to generate, create Code-39 image in Java applications.Recall that a directed acyclic graph has no cycles This important class of graphs simplifies the solution to the shortest-path problem For instance, we do not have to worry about negative-cost cycles because there are no cycles Thus we consider the following problemBarcode Generator In JavaUsing Barcode drawer for Java Control to generate, create barcode image in Java applications.WEIGHTED SINGLE-SOURCE, SHORTEST-PATH PROBLEM FOR ACYCLIC GRAPHS FINDTHE SHORTEST PATH (MEASURED BY TOTAL COST) FROM A DESIGNATED VERTEX S TO EVERY VERTEX IN AN ACYCLIC GRAPH EDGE COSTS AREReading Barcode In JavaUsing Barcode decoder for Java Control to read, scan read, scan image in Java applications.UNRESTRICTED Code-39 Creation In C#.NETUsing Barcode maker for .NET Control to generate, create Code 3/9 image in Visual Studio .NET applications.1551 Topological Sorting Draw USS Code 39 In .NET FrameworkUsing Barcode maker for ASP.NET Control to generate, create USS Code 39 image in ASP.NET applications.Before considering the shortest-path problem, let us examine a related problem: a topological sort A topological sort orders vertices in a directed acyclic graph such that if there is a path from u to v, then v appears after u in the ordering For instance, a graph is typically used to represent the prerequisite requirement for courses at universities An edge (v, w) indicates that course v must be completed before course w may be attempted A topological order of the courses is any sequence that does not violate the prerequisite requiremeritsGenerating Code-39 In .NET FrameworkUsing Barcode encoder for .NET Control to generate, create USS Code 39 image in Visual Studio .NET applications.A topo~ogica~ sort Orders vertices in a d~rected acyclic graph that if there is a path from u to V, then "appears afieruin the ordering A graph that has a cycle havea topological orderCode 39 Extended Drawer In VB.NETUsing Barcode creator for Visual Studio .NET Control to generate, create Code 3/9 image in Visual Studio .NET applications.Graphs and Paths Making UPCA In JavaUsing Barcode creator for Java Control to generate, create UPC-A Supplement 2 image in Java applications.1 / / Single-source negative-weighted shortest-path algorithm & startName ) 3 { 4 vmap::iterator itr = vertexMapfind( startName 1 ; 5 if( itr == vertexMapend( ) j 6 throw Graph~xception( startName + " is not a vertex" ) ; 7 8 clearAll( j ; 9 Vertex *start = (*itr)second; 10 list q; 11 qpush-back( start ) ; start->dist = 0; start->scratch++; 12 while( !qempty( ) ) 13 14 { 15 Vertex *v = qfront ( ) ; qpop-front ( ) ; 16 if( v->scratch++ > 2 * vertexMapsize( j ) 17 throw GraphExceptionj "Negative cycle detected" ) ; 18 19 for( int i = 0; i < v->adjsize( ) ; i++ ) 20 i 21 Edge e = v->adj[ i I ; 22 Vertex *w = edest; 23 double cvw = ecost; 24 if( w->dist > v->dist + cvw ) 25 26 I 27 w->dist = v-zdist + cvw; 28 w->prev = v; 29 / / Enqueue only if not already on the queue 30 if( w->scratch++ % 2 == 0 ) 31 q push-back ( w ) ; 32 else 33 w->scratch++; / / In effect, adds 2 34 } 35 1 36 } 37 }Create Barcode In JavaUsing Barcode maker for Java Control to generate, create bar code image in Java applications.2 void Graph::negative( const string DataMatrix Encoder In JavaUsing Barcode drawer for Java Control to generate, create Data Matrix ECC200 image in Java applications.Figure 1530 A negative-weighted, shortest-path algorithm: Negative edges are allowed GS1-128 Encoder In JavaUsing Barcode maker for Java Control to generate, create EAN / UCC - 13 image in Java applications.Clearly, a topological sort is not possible if a graph has a cycle because, for two vertices v and w on the cycle, there is a path from v to w and w to v Thus any ordering of v and w would contradict one of the two paths A graph may have several topological orders, and in most cases, any legal ordering will doCreating Bar Code In JavaUsing Barcode printer for Java Control to generate, create barcode image in Java applications.Path problems in Acyclic Graphs Making Planet In JavaUsing Barcode generator for Java Control to generate, create USPS PLANET Barcode image in Java applications.Figure 1531 A topological sort The conventions are the same as those in Figure 1522 Barcode Generation In VS .NETUsing Barcode creator for ASP.NET Control to generate, create barcode image in ASP.NET applications.In a simple algorithm for performing a topological sort we first find any vertex v that has no incoming edges Then we print the vertex and logically remove it, along with its edges, from the graph Finally, we apply the same strategy to the rest of the graph More formally, we say that the indegree of a vertex v is the number of incoming edges (u, v) We compute the indegrees of all vertices in the graph In practice, logica11y remove means that we lower the count of incoming edges for each vertex adjacent to v Figure 1531 shows the algorithm applied to an acyclicCreate UCC-128 In .NETUsing Barcode maker for .NET Control to generate, create UCC.EAN - 128 image in VS .NET applications.The indegreeof a vertex is the number of incoming edges A topological sort can be performed in linear time by repeatedly and logically removing vertices that have no incoming edgesEAN13 Creation In .NETUsing Barcode printer for .NET framework Control to generate, create EAN13 image in VS .NET applications.The algorithm produces the correct answer and detects cycles if the graph is not acyclic Painting Bar Code In .NETUsing Barcode drawer for ASP.NET Control to generate, create bar code image in ASP.NET applications.The running time is linear if a queue is used Draw ANSI/AIM Code 39 In Visual C#.NETUsing Barcode drawer for VS .NET Control to generate, create Code 3/9 image in .NET applications.graph The indegree is computed for each vertex Vertex V, has indegree 0, so it is first in the topological order If there were several vertices of indegree 0, we could choose any one of them When V 2 and its edges are removed from the graph, the indegrees of V,,, V,, and V5 are all decremented by 1 Now V has indegree 0, so it is next in the topological order, and V, and V, , have their indegrees lowered The algorithm continues, and the remaining vertices are examined in the order V1, V3, V4 V6,and Vs To reiterate, we do not physically delete edges from the graph; 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