超星电子书《博弈》中文 1 引言 2 棋盘表示 3 走法产生 4 基本搜索技术 5 估值基础 6 研究实例 7 搜索算法改进 8...

JavaScript的帮助文件，经典-JavaScript help files, Classic

网络分析和设计（第二版）――英文版-Network Analysis and Design (2nd Edition)-- English

用C#简单是实现了单词计数程序，C#的初学者可以看看，是个好东东呢-With C# is a simple procedure to achieve the word count, C# can have a look at the beginner, it is a good Dongdong

新编Windows API参考大全.rar-New Windows API Reference Series. Rar

OReilly Java IO,非常不错的书，经典书-OReilly Java IO,a very good book.

c++Primer搞C++必看此书 -c++ Primer engage C++ see this book

很好的关于ＯＲＡＣＬＥ的调优的书 值得参考，现在贡献出来大家共享， 这是第７部分．-good on Oracle optimization book is worth considering, now contribute share, This is the first of seven parts.

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.