资源列表
Huffman-coding
- Huffman编码实现压缩算法，带界面。功能较简单。C++编写。-Huffman coding to achieve compression algorithm, with the interface. Function is relatively simple. C++ writing.
c-tong-xu-lu
- 通讯录实训 c 链接文件 c语言实现简单的通讯录系统 -Training contacts
DirectedGraph
- 改进DFS算法代码示例（判断是否是一个有向无环图）时间复杂度：O(n+e)-Improved DFS algorithm code examples (determine whether there is a directed acyclic graph) Time complexity: O (n+e)
UndirectedGraph
- 遍历一遍，判断图分为几部分（假定为P部分，即图有 P 个连通分量）,对于每一个连通分量，如果无环则只能是树，即：边数＝结点数－1,只要有一个满足边数>结点数－1,原图就有环. -Traverse again, the determination map is divided into several parts (assumed to section P, with P in Figure connected component), for each connected compone
BinaryTreeIterate
- 已知中序和后序遍历，求前序遍历.比较笨的方法是画出来二叉树，然后根据各种遍历不同的特性来求，也可以编程求出.-Known inorder and postorder traversal, seeking preorder traversal. Stupid way is to draw out the binary tree, and then iterate based on a variety of different features to find, but also can be pr
Expression-Evaluator-
- 数据结构，课程设计，表达式计算 c++源码-Data structure curriculum design expression evaluation
GreedyAlgorithm
- Prim,Kruskal,Dijkstra 三个算法的java统一界面实现，可以用同一个界面以及公用一些类与变量，最后还有哈弗曼算法实现，可以作为学习算法的好例子-Prim, Kruskal, Dijkstra algorithm java three unified interface to achieve, you can use the same interface as well as public classes and variables, and finally Huffman a
100programs-in-Python
- 一百个用Python实现的小算法或小程序！绝对实用！如： 【程序56】 题目：画图，学用circle画圆形。 1.程序分析： 2.程序源代码： #include "graphics.h" main() { int driver,mode,i float j=1,k=1 driver=VGA mode=VGAHI initgraph(&driver,&mode,"") setbkcolor(YELLOW) fo
redis2.6.13-modify-intset
- 开源的no-sql的数据库redis的2.6.13版本中，修改了intset的代码，增加了降级功能，可以减少内存的消耗。-Open source no-sql database redis the 2.6.13 version, modified intset code, increasing the degradation function, you can reduce memory consumption.
SegmentTree
- acm的线段树方面的ppt,代码等。适合参加程序设计竞赛的同学-acm aspects of the segment tree ppt, code, etc.. Suitable for students to participate in programming contest
LT8900
- LT8900 2.4G模块程序，收发程序.-LT8900 2.4G module program
data-sources
- 很好的源码可以实现远程监控，远程画面传输-A good source remote monitoring can be achieved