资源列表
StackOperations
- 堆栈操作：初始化堆栈,入栈操作,出栈操作,取栈顶元素。-Stack operations: initialize the stack, the stack operation, and pop operations, to take the top element.
HuffmanTree
- 哈夫曼树：构造哈夫曼树，构造哈夫曼编码。-Huffman tree: Huffman tree structure, constructed Huffman coding.
duobianxing
- 计算机算法实习，使用动态规划法实现多边形游戏，通过调试，里面有详细的解释-The computer algorithm internship polygon game using the dynamic programming method to achieve through the debugger, which are explained in detail
HeapSort
- 堆排序：利用堆进行排序操作，内有详细步骤说明。-Heap Sort: the heap sorting operation, with detailed step-by-step instructions.
正则表达式到有穷自动机转换
- 实现了正则表达式到有穷自动机转换， VC++程序
lvxingfenzhishangxian
- 计算机算法分析实习源码，旅行售货员问题，分枝限界法，通过调试-Computer algorithm analysis internship source, the traveling salesman problem, branch and bound, and through the debugger
FEN
- 分支上限法求解售货员问题，算法分析习题，跟课堂同步实习-Branch ceiling Method salesman problem, algorithm analysis exercises, synchronized with the classroom internship
haiming
- 这个是海明纠错码的C和C++的实现。测试通过，可以正常使用！-This is a Hamming error correction code of C and C++ the realization. Testing through normal use!
C-language-source-code
- 算法导论之排序C语言源代码，几种排序法源码-Introduction to Algorithms Sort C language source code
01bag
- 01 背包问题，c语言实现，计算机算法分析实习，使用的回溯法-01 knapsack problem, the C language, the computer algorithm analysis internship backtracking
2-3-tree
- 2—3树是这样一种树： A．每个非叶子结点都有2个或3个儿子； B．每条众树根到树叶的路径长度相等； C．只有一个根结点的树也是2—3树。 本题目要求从键盘输入以整数序列，建立一棵2-3树。所以我的建树构思为以2-3树的插入操作来进行建树。-2-3 tree is such a tree: A. Each non-leaf node has two or three sons B. Each congregation tree roots to leaves is equal
sampleToValue
- 表达式求值 利用堆栈原理，实现表达式求值 如：3*(3+2)/5#-Expression evaluation Stack principle, the expression is evaluated : 3* (3+2)/5#