lingo最短路0

学习C加加做的实验源程序，希望对大家有用！-Learning to do simple experiments C source code, useful for all of us hope!

windRiver提供的的Zinc5.3完整源码,文档非常齐全，默认状态下可在Windows下用VC编译，支持交叉平台开发，相比高版本的Zinc6.0占用资源要少。-windRiver Zinc5.3 the integrity of the source code, documentation is very complete, default under Windows using VC compiler to support cross-platform development, Compared version of the high-occupancy Zinc6.0 fewer resources.

Servlet for user authenticatiom using session -Servlet for user authenticatiom using session

Dreambox 500S schematic diagram

边角网平差，可以用于导线网平差，是专业课的编程作业，也可用于毕业设计-Corner net adjustment, can be used to wire net adjustment is a specialized courses programming operations, but also can be used for graduation project

自己写的贪吃蛇游戏，体现了完整的面向对象思想-himself wrote the Snake games, reflects the integrity of the object-oriented thinking

这是一篇操作系统的讲义,大家可以好好看一个,希望大家都能有一个好的结果-This is an operating system lectures, we could well see a hope that we can have a good result

一些关于恒流源的资料，那里有电路原理图，用的着的话就拿去吧。-Some information on the constant current source, where the circuit schematic diagram, if used in the case go forward.

可以实现访问一些国内不能访问的站也能访问上-Can realize a number of domestic visits should not visit the station can also visit

本算法使用分治法求解最近点对问题。事先用O(nlogn)时间对x坐标进行排序，使得所有的点是按x坐标从小到大排好序的（x坐标相同时y坐标小的排前），然后取下标小于n/2属于左边的点集PL，取下标大于n/2属于右边的点集PR，即用O(1)时间就可以将规模为n的问题分解为两个规模为n/2的、同类型的子问题。分割完毕之后就可以采用分治法，分别求出PL和PR中的最近点对，最终通过递归实现。-This algorithm uses divide and conquer to solve the problem closest point. Prior to use O (nlogn) time to sort the x coordinate so that all points are based on x coordinates from small to large sorted (x coordinates with the same y coordinates of the small, the top), and then remove the standard is less than n/2 the set of points belonging to the left PL, remove the standard is greater than n/2 set of points belonging to the right of PR, that is to use O (1) time can be the problem size n divided into two size n/2, the same type The sub-problems. Segmentation can be used after completion of sub-rule method, respectively, find the PL and PR in the last points and eventually through the recursion.