Unique Car Software Niche Site – Adsense – Revenue $25 BIN

Why are you selling this site?
Having a clear out of sites, this is a unique site linking to a well known (by word of mouth) bit of freeware. However the software does not have its own site, So i created the site which links to it.. It is maturing well to be picked up on search terms when people look the software up.

Its a fairly simple, easy to use and clean template running wordpress which some clever link adverts – they have made some revenue but nothing to shout about YET – but…

Unique Car Software Niche Site – Adsense – Revenue $ 25 BIN

Algorithm to compute sum of all unique edge pairs of a tree

Given tree is undirected graph. It has n vertices and n-1 edges. The algorithm should compute the sum of all edge pairs. Thus, there are total nC2 or n(n-1)/2 such pairs. The time complexity of the mentioned algorithm is n(n-1)/2. Please suggest an algorithm with better space and time complexity if possible. Below is the java implementation.

import java.util.*;  public class AllPairSumTree {      static long sumAllPairs = 0;      public static void main(String[] args) {         /*          * Total Number of vertices          */         int N = 7;         /*          * Adjacency List          */         LinkedList<Integer>[] adjacencyList = new LinkedList[N];         /*          * Initialize Adjacency List          */         for(int ii=0; ii<N;ii++) {             adjacencyList[ii] = new LinkedList<Integer>();         }         /*          * Weighted Graph Matrix          */         int[][] weightedGraph = new int[N][N];         /*          * Initialize Matrix          */         for(int ii=0;ii<N;ii++) {             for(int jj=0;jj<N;jj++) {                 if(ii == jj) {                     weightedGraph[ii][jj] = 0;                 }else {                     weightedGraph[ii][jj] = Integer.MAX_VALUE;                 }             }         }          /*          * Input Pattern: vertex1,  vertex2, cost          *           * Total Vertex: N, Total Edges: N-1 (Tree, Undirected Graph)          */         int[] inputGraph = { 1, 2, 1,                                 2, 3, 2,                                 3, 4, 3,                                 3, 5, 4,                                 5, 6, 6,                                 5, 7, 5};          /*          * Assign  Adjacency List  and Matrix with input Graph          */         for(int ii=0; ii<N-1; ii++) {             int vertex1 = inputGraph[ii*3 + 0] - 1;             int vertex2 = inputGraph[ii*3 + 1] - 1;             int cost = inputGraph[ii*3 + 2];              adjacencyList[vertex1].add(vertex2);             adjacencyList[vertex2].add(vertex1);    //bidirectional edge              weightedGraph[vertex1][vertex2] = cost;             weightedGraph[vertex2][vertex1] = cost; //bidirectional edge         }          sumAllPairs = 0;         int currentVertex = 0;         LinkedHashSet<Integer> visitedSet = new LinkedHashSet<Integer>(N);         int lastVisitedVertex = -1;         allPairSum(weightedGraph, adjacencyList, currentVertex, visitedSet, lastVisitedVertex);          System.out.println(sumAllPairs);     }      /*      * Say graph has vertices 1,2,3,4,5,6,7      *       * allPairSum() will compute sum of edges like this:      *              21 + (31+32) + (41+42+43) + (51+52+53+54) + (61+62+63+64+65)+(71+72+73+74+75+76)      *              where ij represents edge from vertex i to vertex j      *       * Time Complexity:      *      N(N-1)/2 or Combination(N,2)[![enter image description here][1]][1]      */     private static void allPairSum(int[][] weightedGraph, LinkedList<Integer>[] adjacencyList, int currentVertex, LinkedHashSet<Integer> visitedSet, int lastVisitedVertex) {          for(Integer visitedVer : visitedSet) {             int cost = weightedGraph[visitedVer][lastVisitedVertex] +  weightedGraph[lastVisitedVertex][currentVertex];             sumAllPairs += cost;             weightedGraph[visitedVer][currentVertex] = cost;             weightedGraph[currentVertex][visitedVer] = cost;         }          visitedSet.add(currentVertex);          for(Integer neighbourVert : adjacencyList[currentVertex]) {             if(neighbourVert != lastVisitedVertex) {                 /*                  *      neighbourVert becomes currentVertex                  *      currentVertex becomes lastVisitedVertex                  */                 allPairSum(weightedGraph, adjacencyList, neighbourVert, visitedSet, currentVertex);             }         }     } } 

Sample tree with 7 vertices and 6 edges

Create Unique domain 40 High Pr Dofollow Banklinks Blogcomments PR2 to PR7 for $2

★★★★ Thanks for View my New Gig ★★★★ Improving your SEO ranking is easier than ever. PR7 To PR3 All Dofollow Blog Comment Low obl links On Actual Page One of the most Powerful link building strategy is to creating High authority dofollow blogs Properties PR 7 to PR 2 as a part of a biggest strategy. This takes works and efforts, Those links are LONG-TERM, & hold a TON of weight. Key Features: > All Niches Accepted > Unlimited URLs and KEYWORDS accepted. > Total handmade manually Work. > 100% Satisfaction Guaranteed. > DoFollow Links > Not Use Any Automated Software > No spam work > 24/7 Customer Support > After Work full summary detail with high profile Excel report > Unlimited URLs and KEYWORDS accepted. > High-Quality Relevant Content > On Time Delivery This will in turn create backlinks to high-ranking websites so that you improve your SEO rankings. Before you know it, you’ll have more traffic to your website – and generate more business. It allows you to stomp out your competition, and with our help, you don’t have to fuss with all of the search engine optimization tactics on your own. It saves you time and gets you the desired results.

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I provide You 100 Unique Votes for $20

Hello I am seo and smm expert. I have 3 years experience for this job. I have a big team and expert worker. I will play online contest. I will give you unlimited online contest Votes.Such as,- Real IP votes, Signup votes, mail Confirmation votes etc… If you need to my service please contact me. Thanks

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High Quality Unique Business Card design in 24 hours for $10

Hello Buyers! Welcome to my Gig Description. I will design nowadays Smart and Professional Business Card for your Business. My design will be 100% real and full copy right protected with you. Give me your Company Logo or I will design your logo but extra cost $ 30 in all Packages. Silver Package: ($ 10)1. Business Card (QR Code or Bar Code) 1. Single or Double Side 2. Letterhead or Envelopes 3. Revision (1,2) Gold Package: ($ 15) 1. Business Card (QR Code or Bar Code) 2. Letterhead + Envelopes 3. Single or Double Side 4. Print Ready (JPEG + PDF) 5. Source File (AI + PSD) 6. Revision (Unlimited) Diamond Package: ($ 30) 1. Professional Business Card (QR Code or Bar Code) 2. Letterhead + Envelope + Cover 3. Single or Double Side 4. JPEG + PDF Formats 5. Source File (AI + PSD) 6. Revision (Unlimited) Card Dimension: · 3.5″ x 2″ or any other dimension you’ll need. · The business card can be horizontal or vertical. Please Contact me if you need any clarification, questions and queries. So why are you waiting, Order Your Professional Business Card and I will happy to work with you. Thank You

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Unique values for x-axis in a number chart

I have a numbers table with a bunch of values in a column, and a count of the respective value in another column. If I am plotting this in a chart, how can I make sure that the values in the one of the column (x-axis) is used uniquely?

Here is a simplistic example of what I have:

enter image description here

How can I make it so that the x-axis only show Apples and Bananas as to bars, and the total count on the y-axis? Do I need to create a separate table to do that calculation and plot that table? Is there any better/elegant way of handling this?

Keychain unique user

I want to be able to store and retrieve a unique user throughout the app. Multiple accounts are not used. The code I have written is stated below.

This is my first attempt using KeyChain. The test succeeds. I am wondering if anything if the code can be improved. Security issues perhaps?

import XCTest  class KeyChain {     struct User {         let identifier: Int64         let password: String     }      static func save(user: User) -> Bool {         let identifier = Data(from: user.identifier)         let password = user.password.data(using: .utf8)!         let query = [kSecClass as String : kSecClassGenericPassword as String,                      kSecAttrService as String : "comapp",                      kSecAttrAccount as String : identifier,                      kSecValueData as String : password]             as [String : Any]          let deleteStatus = SecItemDelete(query as CFDictionary)          if deleteStatus == noErr || deleteStatus == errSecItemNotFound {             return SecItemAdd(query as CFDictionary, nil) == noErr         }          return false     }      static func retrieveUser() -> User? {         let query = [kSecClass as String : kSecClassGenericPassword,                      kSecAttrService as String : "comapp",                      kSecReturnAttributes as String : kCFBooleanTrue!,                      kSecReturnData as String: kCFBooleanTrue!]             as [String : Any]          var result: AnyObject? = nil         let status = SecItemCopyMatching(query as CFDictionary, &result)          if status == noErr,             let dict = result as? [String: Any],             let passwordData = dict[String(kSecValueData)] as? Data,             let password = String(data: passwordData, encoding: .utf8),             let identifier = (dict[String(kSecAttrAccount)] as? Data)?.to(type: Int64.self) {              return User(identifier: identifier, password: password)         } else {             return nil         }     } }  private extension Data {     init<T>(from value: T) {         var value = value          self.init(buffer: UnsafeBufferPointer(start: &value, count: 1))     }      func to<T>(type: T.Type) -> T {         withUnsafeBytes { $  0.load(as: T.self) }     } }   class SecureStoreTests: XCTestCase {     func testUser() {         let user = KeyChain.User(identifier: 200, password: "somePassword")          let didAddedUser = KeyChain.save(user: user)          XCTAssertTrue(didAddedUser)          let retrievedUser = KeyChain.retrieveUser()!          XCTAssert(user.identifier == retrievedUser.identifier)         XCTAssert(user.password == retrievedUser.password)     } } 

Erro ao modificar valor de uma Unique Key no ASP .NET Core

Quando eu tento fazer o Update dos dados do Aluno o seguinte erro acontece: “Duplicate entry ‘teste@gmail.com’ for key ‘AlternateKey_EmailConta'”

O campo EmailConta é uma Unique Key.

Isso acontece com todos os campos Unique Key do banco então obviamente é algo que eu não estou sabendo fazer direito.

Se alguém puder me ajudar fico muito agradecido 😀

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Unique ID for 2 multiple search box

Unfortunately as I concluded, If you want to work with two search-box in a sharepoint page (for example one in the layout page and other in the master page/ or simply add two sharepoint search box web parts to a webpart zone. all of the search boxes have a div element with same id (id="SearchBox") and this make some misfunctions (for example all the search box automaticly get the value which you type in one of them). What is the solution for this situation? Is it logically wrong to add more than one search-box in a page? Thanks in advance.