C Language Advanced (East Grand Ming Ling very good) for those who have c basis

TCP-IP相关技术-TCP-IP related technology

年轻工程师是如何锻炼成“高手”-young engineers is how to exercise as a "master"

c#wendan 其他人不需帐号就可自由下载此源码 c#的一些入门资料 -c#wendan 其他人不需帐号就可自由下载此源码 c#的一些入门资料

计算式求值栈操作-formula evaluates stack operation

< linux系统分析与高级编程> 6-10章- Chapter 6-10

Linux kernal 核心中文手册-Chinese Linux core manual Geekondu

VC++技术内幕源码-VC insider source technology

关于数字电路设计仿真设计的仿真设计文件的编写教程-on digital circuit design simulation design of the simulation design documents prepared Directory

网友学习笔记.rar-Netizen study notes. Rar

背包问题动态规划算法JAVA 给定n种物品和一背包。物品i的重量是wi，其价值为vi，背包的容量为c。问应如何选择装入背包中的物品，使得装入背包中物品的总价值最大？ 在选择装入背包的物品时，对每种物品i只有两种选择，即装入背包或不装入背包。不能将物品i装入背包多次，也不能只装入部分的物品i。因此，该问题称为0-1背包问题。-Knapsack problem dynamic programming algorithm for a given n species JAVA goods and one backpack. Item i is the weight of wi, the value for vi, knapsack capacity c. Question should be how to choose the items into the backpack, the backpack into the total value of the largest items? In the selection of the items into the backpack, for each of the items i have only two choices, that is loaded or not loaded backpack backpack. Items should not be many times i load backpack can not only part of the goods loaded i. Therefore, the problem known as the 0-1 knapsack problem.