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SR-tree

This is an implementation of double-array structure for representing trie, as proposed by Junichi A

This is an implementation of double-array structure for representing trie, as proposed by Junichi Aoe [1]. Trie is a kind of digital search tree, an efficient indexing method with O(1) time complexity for searching. Comparably as efficient as hashing, trie also provides flexibility on incremental matching and key spelling manipulation. This makes it ideal for lexical analyzers, as well as spelling dictionaries. See the details of the implementation at [2]: http://linux.thai.net/~thep/datrie/datrie.html Historically, this was first implemented as C++ classes in a library called midatrie [2], but later simplified and rewritten from scratch in C.

標簽： implementation double-array representing structure

上傳時間： 2013-12-10

上傳用戶：shinesyh
spoj dtree. Count the number of distinct elements in a given range. Usage is done using fenwick tree

spoj dtree. Count the number of distinct elements in a given range. Usage is done using fenwick trees. Offline algorithm for queries

標簽： distinct elements fenwick number

上傳時間： 2014-01-20

上傳用戶：lhw888
自己編的

自己編的，在VB中實現普里姆最小生成樹算法的代碼，有圖的，邊的權值是隨機生成的，5個節點，點“generate tree”，就會顯示出最小樹。

標簽：

上傳時間： 2017-04-09

上傳用戶：Andy123456
一個簡單的Ext例子

一個簡單的Ext例子，里面有Ext.grid.GridPanel和Ext.tree.TreePanel等Ext常用控件的使用，以及Ext和后臺的交互

標簽： Ext

上傳時間： 2013-12-23

上傳用戶：qweqweqwe
伸展樹

伸展樹，基本數據結構，The tree is drawn in such a way that both of the edges down from a node are the same length. This length is the minimum such that the two subtrees are separated by at least two blanks.

標簽： 樹

上傳時間： 2017-05-07

上傳用戶：JIUSHICHEN
ZigBee路由協議指的是ZigBee規范中規定的與路由相關的功能和算法部分

ZigBee路由協議指的是ZigBee規范中規定的與路由相關的功能和算法部分，主要包括不同網絡拓撲結構下ZigBee協議數據單元的路由方式、路由發現和路由維護等內容。為了達到低成本、低功耗、可靠性高等設計目標，ZigBee網絡采用了Cluster-Tree與AODV路由相結合的路由算法。該包實現了ZIGBEE路由算法在NS2下的模擬

標簽： ZigBee 路由協議路由算法

上傳時間： 2017-05-14

上傳用戶：WMC_geophy
一個實現struts+spring+hebernate的例子

一個實現struts+spring+hebernate的例子，外加一個JavaScript實現的tree的都OK了。

標簽： hebernate struts spring

上傳時間： 2013-11-25

上傳用戶：skhlm
Huffman codes 1.In telecommunication, how do we represent a set of messages, each with an access fr

Huffman codes 1.In telecommunication, how do we represent a set of messages, each with an access frequency, by a sequence of 0 s and 1 s? 2.To minimize the transmission and decoding costs, we may use short strings to represent more frequently used messages. 3.This problem can be solved by using an extended binary tree which is used in the 2-way merging problem. the sourse code is to encode and decode the Huffman.

標簽： telecommunication represent messages Huffman

上傳時間： 2013-12-22

上傳用戶：ztj182002
Traveling Salesperson Problem Our branch-and-strategy splits a branch and bound solution into two

Traveling Salesperson Problem Our branch-and-strategy splits a branch and bound solution into two groups: one group including a particular arc and the other excluding this arc. 1.Each splitting incurs a lower bound and we shall traverse the searching tree with the "lower" lower bound. 2.If a constant subtracted from any row or any column of the cost matrix, an optimal solution does not change.

標簽： branch-and-strategy Salesperson Traveling solution

上傳時間： 2013-12-29

上傳用戶：璇珠官人
) Compression using huffman code -with a number of bits k per code word -provide huffma

) Compression using huffman code -with a number of bits k per code word -provide huffman table Huffman coding is optimal for a symbol-by-symbol coding with a known input probability distribution.This technique uses a variable-length code table for encoding a source symbol. The table is derived in a particular way based on the estimated probability of occurrence for each possible value of the source symbol .Huffman coding uses a specific method for representing each symbol, resulting in a prefix code that expresses the most common characters using shorter strings of bits than those used for less common source symbols.The Huffman coding is a procedure to generate a binary code tree.

標簽： code Compression huffman provide

上傳時間： 2017-05-30

上傳用戶：yuchunhai1990