Maze

求出任意两个序列的最长公共子序列,使用了动态规划的方法-Obtained the longest sequence of any two common sub-sequences, the use of dynamic programming methods

这是我本人编写的一个32位定点小数运算的函数库。对于没有浮点运算器的场合，可以提供进行小数运算的一种途径-This is my prepared in a 32-bit fixed point decimal computing functions. For no floating point unit occasions, it can provide for fractional arithmetic, as a way to

数据库实例，提供了汉字首字母转换的详细信息，相信大家能得到。

这是一个用C语言编写的哈夫曼编/译码器程序，与严蔚敏版数据结构习题的5.2相对应啊-This is a C language prepared by the Huffman encoder/decoder procedures, and Yan Wei-min data structure Exercise version 5.2 should be relatively ah

参加运动会的 个学校编号为 。比赛分成 个男子项目和 个女子项目，项目编号分别为 和 。由于各项目参加人数差别较大，有些项目取前五名，得分顺序为7，5，3，2，1；还有些项目只取前三名，得分顺序为5，3，2。写一个统计程序产生各种成绩单和得分报表。 2、要求产生各学校的成绩单，内容包括各校所取得的每项成绩的项目号、名次（成绩）、姓名和得分；产生团体总分报表，内容包括校号、男子团体总分、女子团体总分和团体总分。 3、测试数据：对于 ， ， ，编号为奇数的项目取前五名，编号为偶数的项目取前三名，设计一组实例数据。 (The school serial number participating in Games is. Competition divides into men s event and women s event , the project serial number parts for the sum. The difference is bigger since every project participates in a number , some projects choose the first five , score order is 7 , 5 , 3 , 2 , 1 Still have some of projects taking the first three places , score only being 5 , 3 , 2 in proper order. Write a form for report counting procedure producing the various school report card and score. 2, demands the school report card producing every school , content includes every achievement project number , position in a name list (achievement) , full name and score per got by school Produce the group total points form for report , content including the school number , male person group total points , woman group total points and group total points. 3, testing data: Be that the odd number project chooses the first five to the serial number, the serial number is that the even number project c)

单源最短路径算法的C语言实现，刚学算法时写的，多多指教

根据数据结构教材，利用循环队列，非常不错-based data structure materials, the use of recycling cohort, very good

1 2 0.0403 1 3 0.4747 1 4 0.7015 1 5 0.7936 1 6 0.2973 1 7 0.6139 1 8 0.4494 1 9 0.3155 1 10 0.8558 1 11 0.272 1 12 0.1045 1 13 0.6548 1 14 0.9271 1 15 0.3229 2 1 0.2182 2 3 0.9329 2 4 0.9521 2 5 0.8128 2 6 0.4044 2 7 0.6619 2 8 0.6596 2 9 0.3007 2 10 0.7244 2 11 0.2316 2 12 0.01 2 13 0.915 2 14 0.0878 2 15 0.0984 3 1 0.967 3 2 0.8945 3 4 0.749 3 5 0.9038 3 6 0.3022 3 7 0.2 3 8 0.7532 3 9 0.042 3 10 0.1991 3 11 0.8995 3 12 0.0592 3 13 0.4332 3 14 0.3324 3 15 0.17 4 1 0.434 4 2 0.3857 4 3 0.5991 4 5 0.5404 4 6 0.7573 4 7 0.96 4 8 0.8047 4 9 0.5279 4 10 0.1573 4 11 0.9087 4 12 0.3227 4 13 0.2898 4 14 0.5262

资源描述数据结构第三章课后题二叉树，对于初学数据结构有很大的帮助

别人的VC操作Access数据库都是用的GUI,让VC手生的看了迷糊,不得要领,这次我封装一个纯C++操作Access2000库的源代,主要封装了一个访问Access数据库的类,非常简单.-Someone else" s VC action Access databases are used in GUI, so that VC looked rusty" s confused, but to no avail, this time I packaged a pure C++ source operating Access2000 database generation, mainly encapsulates an Access database, access to classes, is very simple .