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lvxingshang
一个经典的旅行商问题源码，不是最优解，只是一种 思路(A classic traveling salesman problem source is not the optimal solution, just a thought)
- 2012-11-09 01:33:02下载
- 积分：1
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maopao_youhua
冒泡优化：
如果一个序列是int n[]={1,2,3,4,5,6,7,8,9} , 用正常的冒泡排序需要排8次才行，优化之后1次就好，也就是说序列越接近于正常序列，改进之后的冒泡排序的次数就越少，这样会给一个冒泡排序算法带了很大的效率。
思想：添加一个boolean变量用来判断冒泡是否是已经排好了顺序，如果boolean的值为false，说明是已经排好了，如果boolean的值true，说明没有排好，继续排。(If a sequence is int n [] = {, 1,2,3,4,5,6,7,8,9} need to row 8 times the job after optimization times like normal bubble sort, but alsomeans that the sequence is more close to the normal sequence, improved bubble sort, the less the number, which would give a bubble sort algorithm with a great deal of efficiency.
Idea: add a boolean variable used to determine the bubble whether it is already lined up the order, if the boolean is false, indicating already lined up, if the boolean value of true, did not line up, continue to row.)
- 2012-04-18 00:33:45下载
- 积分：1
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Monkeys-and-peach
Monkeys and peach,Monkeys and peach(Monkeys and peach)
- 2013-09-01 22:06:42下载
- 积分：1
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Joseph
说明： 约瑟夫问题，四种解法，分别是静态链表，顺序表两种和循环链表(Joseph problems, the four solution, namely a static list, order form and the cycle of two lists)
- 2010-04-23 08:20:53下载
- 积分：1
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xiao-yuan-dao-you
说明： 设计济南大学校园平面图，所含的景点不少于10个。以图中顶点表示校内各景点，存放景点名称、代号、简介等信息；以边表示路径，存放路径长度等相关信息。
为来访客人提供图中任意景点相关信息的查询。
为来访客人提供图中任意景点的问路查询，即查询任意两个景点之间的一条最短的简单路径。
选作内容：
求校园的关节点。
提供图中任意景点查询，即求任意两点间的所有路径。
校园导游图的景点和道路的修改扩充功能。
扩充道路信息，如道路类别，沿途景色等级，一直可按客人所需分别查询人行路径或车行路径或观景路径。(Design jinan university campus plan, contains at least 10 spots in each vertex said in scenic spots, deposit attractions at the information such as the code name introduction On the path, deposit path length said related information
)
- 2011-03-21 19:54:06下载
- 积分：1
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Huffman-Tree
使用三叉链表实现的哈夫曼树,统计inputfile1.txt中各字符的出现频率，并据此构造Huffman树，编制Huffman 码；根据已经得到的编码，对01形式的编码段进行译码。
具体的要求：
1．将给定字符文件编码，生成编码，输出每个字符出现的次数和编码；
2．将给定编码文件译码，生成字符，输出编码及其对应字符。
(Emergence of the frequency of each character in the trigeminal lists using the Huffman tree, statistics inputfile1.txt, and accordingly the Huffman tree structure, preparation of Huffman code according to the coding has been. 01 to form the code segment for decoding.
Specific requirements:
1 coding the given character file, generating the encoding, and outputting the number and encoding of each character
2 will be given the encoding file decoding, generating the character, the output code and its corresponding character.)
- 2015-05-22 15:37:03下载
- 积分：1
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Czhihen
关于C语言的一些指针的处理算法，能够使得初学者对C语言指针有自己的了解(Some pointers on C language processing algorithms, it is possible to make a beginner C language pointer has its own understanding)
- 2016-06-17 11:29:59下载
- 积分：1
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d
有助于c++的初学者在平日的学习中更好的学习(Contribute to better learning c++ beginners learning on weekdays)
- 2013-11-29 14:32:18下载
- 积分：1
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tcpip
以太网首部的一个c语言程序，代码功能是对设备要传送的帧进行排队(Ethernet header c language program code function is queued frames to be transferred to the device)
- 2013-03-26 11:49:23下载
- 积分：1
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B_Tree
B-树的创建、插入、删除等一系列操作！(B-tree creation, insert, delete a series of manipulations!)
- 2020-10-18 18:17:26下载
- 积分：1
