资源列表
yiyuan
- 医院模拟:经典堆栈队列应用；VC++控制台程序；小弟上数据结构是作品.-hospital simulation : classic stack Queue application; VC console; Finding data on the structure of work.
ExpressionCalculator
- 用VC写的一个根据输入的数学表达式字符串求出结果，将后缀转为中缀，类似一个计算器，唯一遗憾的是只是整形之间的+—*/。-VC was based on the importation of a mathematical expression string of results obtained, which is made up to the suffix similar to a calculator, the only regret is the only plastic -* /.
huowuyunshudeyichuansuanfa
- #include \"stdio.h\" #include \"math.h\" #include \"malloc.h\" #include \"string.h\" #define m 6 #define n 29 #define p 0.5-#include \"stdio.h\" #include \"math.h\" #include \"malloc.h\" #include \"string.h\" #define m 6 #define n 29 #def
FLOYDsuanfa
- #include \"stdio.h\" #include \"math.h\" #include \"malloc.h\" #include \"string.h\" #define m 6 #define n 29 #define p 0.5-#include \"stdio.h\" #include \"math.h\" #include \"malloc.h\" #include \"string.h\" #define m 6 #define n 29 #def
juigon
- 九宫问题(八数码)求解过程动态演示 用VC＋＋编程实现，可以直观地看到演示效果！-JiuGongTu (Digital 8) solution process dynamic demonstration with VC + + programming, visual effects demo to see!
Tansuo
- 数据结构－－－堆栈的使用。蓝色小方块在一个自己添加的红色迷宫中找到一个出口。利用数据结构堆栈的特性来。-data structure --- stack use. Blue small chunks themselves in a maze of red added to find an export. Stack data structure used to the characteristics.
accept
- 数字运算，判断一个数是否接近素数 A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in
accepted1
- 解决时钟问题，acm竞赛题 A weird clock marked from 0 to 59 has only a minute hand. It won t move until a special coin is thrown into its box. There are different kinds of coins as your options. However once you make your choice, you cannot use any other kind.
Estimate
- 理解并掌握分析回溯算法效率的方法。 使用Estimate算法就8皇后问题给出效率估算 -understanding and mastery of backtracking algorithm efficiency analysis method. Use Estimate on 8 Queen's algorithm is efficient estimation problem
Strassen_8
- Strassn关于两个矩阵相乘的算法，同过分治原理把两个n*n的矩阵阶各分解成四个n/2*n/2阶的矩阵，当分解出来的矩阵阶数等于2时，求借各个小矩阵，若阶数大与2，就递归的调用前面方法，直到分解成2*2的子矩阵为止。-Strassn on two matrix multiplication, the algorithm, with the governing principle over two n * n matrix of the band decomposed into 4 n / 2
graphic_algorithm
- 每对节点间最短路径 Floyd-Warshall 算法 D[i,j]表示从i到j的最短距离； P[i,j]表示从i到j的最短路径上j 的父节点-between each pair of nodes Shortest Path Floyd-Warshall algorithm D [i, j] said from i to j is the shortest distance; P [i, j] said from i to j the shortest path
codeandencode
- 对一篇文章编码和解码，采用huffman 树的方法-an article on encoding and decoding, using the Huffman tree