 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
how to generate a barcode using asp.net c# (u,v) E in Software
(u,v) E QR Encoder In None Using Barcode creation for Software Control to generate, create QR Code ISO/IEC18004 image in Software applications. QR Code ISO/IEC18004 Reader In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. min {dist(u, i 1) + (u, v)} Generate Quick Response Code In C#.NET Using Barcode creation for .NET Control to generate, create QR Code image in VS .NET applications. QR Code Printer In VS .NET Using Barcode generator for ASP.NET Control to generate, create QRCode image in ASP.NET applications. Allpairs shortest paths
QR Code ISO/IEC18004 Generation In VS .NET Using Barcode generator for .NET framework Control to generate, create QRCode image in Visual Studio .NET applications. Generating Quick Response Code In VB.NET Using Barcode drawer for .NET Control to generate, create Denso QR Bar Code image in VS .NET applications. What if we want to nd the shortest path not just between s and t but between all pairs of vertices One approach would be to execute our general shortestpath algorithm from Section 461 (since there may be negative edges) V  times, once for each starting node The total running time would then be O(V  2 E) We ll now see a better alternative, the O(V  3 ) dynamic programmingbased FloydWarshall algorithm Is there is a good subproblem for computing distances between all pairs of vertices in a graph Simply solving the problem for more and more pairs or starting points is unhelpful, 172 Encoding Code 128 Code Set A In None Using Barcode printer for Software Control to generate, create Code 128C image in Software applications. GS1 128 Maker In None Using Barcode generator for Software Control to generate, create UCC128 image in Software applications. because it leads right back to the O(V  2 E) algorithm One idea comes to mind: the shortest path u w 1 wl v between u and v uses some number of intermediate nodes possibly none Suppose we disallow intermediate nodes altogether Then we can solve allpairs shortest paths at once: the shortest path from u to v is simply the direct edge (u, v), if it exists What if we now gradually expand the set of permissible intermediate nodes We can do this one node at a time, updating the shortest path lengths at each stage Eventually this set grows to all of V , at which point all vertices are allowed to be on all paths, and we have found the true shortest paths between vertices of the graph! More concretely, number the vertices in V as {1, 2, , n}, and let dist(i, j, k) denote the length of the shortest path from i to j in which only nodes {1, 2, , k} can be used as intermediates Initially, dist(i, j, 0) is the length of the direct edge between i and j, if it exists, and is otherwise What happens when we expand the intermediate set to include an extra node k We must reexamine all pairs i, j and check whether using k as an intermediate point gives us a shorter path from i to j But this is easy: a shortest path from i to j that uses k along with possibly other lowernumbered intermediate nodes goes through k just once (why because we assume that there are no negative cycles) And we have already calculated the length of the shortest path from i to k and from k to j using only lowernumbered vertices: Painting Bar Code In None Using Barcode printer for Software Control to generate, create barcode image in Software applications. Barcode Drawer In None Using Barcode generator for Software Control to generate, create barcode image in Software applications. dist(i, k, k 1) i dist(i, j, k 1) European Article Number 13 Maker In None Using Barcode encoder for Software Control to generate, create EAN13 image in Software applications. Data Matrix ECC200 Creator In None Using Barcode maker for Software Control to generate, create Data Matrix 2d barcode image in Software applications. k dist(k, j, k 1) j
Draw Identcode In None Using Barcode creation for Software Control to generate, create Identcode image in Software applications. Data Matrix 2d Barcode Drawer In C# Using Barcode maker for Visual Studio .NET Control to generate, create Data Matrix image in .NET framework applications. Thus, using k gives us a shorter path from i to j if and only if dist(i, k, k 1) + dist(k, j, k 1) < dist(i, j, k 1), in which case dist(i, j, k) should be updated accordingly Here is the FloydWarshall algorithm and as you can see, it takes O(V  3 ) time for i = 1 to n: for j = 1 to n: dist(i, j, 0) = for all (i, j) E: dist(i, j, 0) = (i, j) for k = 1 to n: for i = 1 to n: for j = 1 to n: dist(i, j, k) = min{dist(i, k, k 1) + dist(k, j, k 1), dist(i, j, k 1)} UCC  12 Generation In None Using Barcode printer for Office Excel Control to generate, create UPCA image in Excel applications. Painting UPC  13 In ObjectiveC Using Barcode printer for iPad Control to generate, create GS1  13 image in iPad applications. The traveling salesman problem
EAN / UCC  14 Generator In .NET Using Barcode creator for VS .NET Control to generate, create EAN / UCC  14 image in VS .NET applications. Decoding Code 128 Code Set A In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. A traveling salesman is getting ready for a big sales tour Starting at his hometown, suitcase in hand, he will conduct a journey in which each of his target cities is visited exactly once 173 Barcode Creator In None Using Barcode creator for Office Excel Control to generate, create bar code image in Office Excel applications. Code 128 Code Set B Decoder In C#.NET Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications. Figure 69 The optimal traveling salesman tour has length 10
2 1 3 before he returns home Given the pairwise distances between cities, what is the best order in which to visit them, so as to minimize the overall distance traveled Denote the cities by 1, , n, the salesman s hometown being 1, and let D = (d ij ) be the matrix of intercity distances The goal is to design a tour that starts and ends at 1, includes all other cities exactly once, and has minimum total length Figure 69 shows an example involving ve cities Can you spot the optimal tour Even in this tiny example, it is tricky for a human to nd the solution; imagine what happens when hundreds of cities are involved It turns out this problem is also dif cult for computers In fact, the traveling salesman problem (TSP) is one of the most notorious computational tasks There is a long history of attempts at solving it, a long saga of failures and partial successes, and along the way, major advances in algorithms and complexity theory The most basic piece of bad news about the TSP, which we will better understand in 8, is that it is highly unlikely to be solvable in polynomial time How long does it take, then Well, the bruteforce approach is to evaluate every possible tour and return the best one Since there are (n 1)! possibilities, this strategy takes O(n!) time We will now see that dynamic programming yields a much faster solution, though not a polynomial one What is the appropriate subproblem for the TSP Subproblems refer to partial solutions, and in this case the most obvious partial solution is the initial portion of a tour Suppose we have started at city 1 as required, have visited a few cities, and are now in city j What information do we need in order to extend this partial tour We certainly need to know j, since this will determine which cities are most convenient to visit next And we also need to know all the cities visited so far, so that we don t repeat any of them Here, then, is an appropriate subproblem For a subset of cities S {1, 2, , n} that includes 1, and j S, let C(S, j) be the length of the shortest path visiting each node in S exactly once, starting at 1 and ending at j When S > 1, we de ne C(S, 1) = since the path cannot both start and end at 1 Now, let s express C(S, j) in terms of smaller subproblems We need to start at 1 and end at j; what should we pick as the secondtolast city It has to be some i S, so the overall path length is the distance from 1 to i, namely, C(S {j}, i), plus the length of the nal edge, 174

