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西门子与计算机通讯程序包括源代码 程序说明 控件 及控件使用手册 使用西门子prodave 控件 这是第一版-Siemens and computer communication procedures, including source code description of the procedures for the use of manual controls and controls the use of Siemens controls prodave This is the first edition of

零件切割问题： 给定一块宽度为W的矩形板，矩形板的高度不受限制。现需要从板上分别切割出n个高度为hi，宽度为wi的矩形零件。切割的规则是零件的高度方向与矩形板的高度方向保持一致。问如何切割使得所使用的矩形板的高度h最小？ 任给一个输入实例，能输出切割所需要的实际高度并能用图形演示切割的过程-cutting parts : a given width of the rectangular plate W, the height of rectangular plate unrestricted. Need from the board is cutting out respectively n height of the hi, wi width of the rectangular parts. Cutting parts to the rules of the height of the rectangular plate with a high degree of consistent direction. Asked how cutting makes use of the rectangular plate height h minimum? As an input for example, will cut output by the needs of the high and can be used graphic demonstration of the cutting proce

fifo和lru页面置换算法-fifo and LRU algorithm replacement pages

Realiza inventarios no terminado

不带头结点的链式堆栈类,1、 定义并实现不带头结点的链式堆栈类，除构造和析构函数外，成员函数还有入栈、出栈、取栈顶元素、判断堆栈是否非空，头文件为“nohead_LinStack.h”。-node does not take the lead in the category of chain stack, 1, and the definition does not take the lead in realizing the chain nodes stack category, In addition to construction and destructor function, the members also function stack, stack up and stack elements for judging whether non-empty stack, The first document as "nohead_LinStack.h."

A*算法只要求产生问题的全部状态空间的部分结点及关系,就可以求解问题了,搜索效率较高。当然由于A*算法中没有回溯，因此某些情况下可能会找不到正确的解过程。 -A* algorithm only requires a problem all part of the state space and the relationship between nodes, we can solve the problem, the search more efficient. Of course, as a result of A* algorithm is not retroactive, so some cases may be unable to find the correct solution of the process.

这是学习和；

最短路径查询,用的是ESRI公司arcobject组件开发的,已经作成dll,用起来非常的方便-Shortest Path inquiries, the company is ESRI arcobject development of the components, make dll with it is very convenient

程序读取所有；

在0和1情况下进行背包问题解决，涉及多个物体的重量和价值，求得最大价值量以及分配方案。-At 0 and 1 backpack case for problem-solving, involving more than the weight of objects and values, as well as to achieve maximum value distribution program.