linux-mail

一个初级程序文档，按照这个做，你很快就能摆脱菜鸟哦(A primary program documentation, according to this so quickly you can get rid of rookie Oh)

UML的重要资料，我看后获益匪浅，大家可以下载来看看，随便下载。(UML important information, I look after the benefit, everyone can download to look at, just download.)

2.1寻找背包数组及遍历工具写法图文演示！2.1寻找背包数组及遍历工具写法图文演示！(2.1 Array and find backpack traverse graphic demonstration of writing instrument! 2.1 Array and find backpack traverse graphic demonstration of writing instrument!)

这是一个好用的频偏估计算法的matlab仿真程序,含噪脉冲信号进行相关检测,采用波束成形技术的BER计算。( This is a useful frequency estimation algorithm matlab simulation program, Noisy pulse correlation detection signal, By applying the beam forming technology of BE.)

关于自适应旁瓣对消的文章，看看有无用处。(Adaptive sidelobe cancellation of the article to see whether the use.)

papers helpful for understanding design of induction motors

说明：  基于云模型的数字影像质量评估，专业文章，测绘相关，呵呵~~~~~~(Cloud Model-based digital video quality assessment, professional article, mapping-related, huh, huh ~~~~~~)

软件工程课程设计报告，飞机票售票系统，相信对你们会有帮助的(Plane ticket ticketing system software engineering curriculum design report in detail, believe that will be helpful to you )

K3 table Struce PPT trening

1、 创建图类，存储结构使用邻接矩阵。 2、 输入图的节点数n（不超过10个）、边数m，节点分别用1-n代表。 3、 采用“起始节点，终止节点，权值”输入图的m条边，创建图。 4、 输出从节点1开始的BFS遍历，在遍历过程中，如有多个可以选择的节点，则优先选择编号较小的节点。 5、 输出从节点1开始的DFS遍历，在遍历过程中，如有多个可以选择的节点，则优先选择编号较小的节点。 6、 输出从第1节点到第n节点最短路径的长度，如果没有路经，输出0。(1. Create graph class and use adjacency matrix in storage structure. 2. The number of nodes n (no more than 10) and edge m in the input graph are represented by 1-n, respectively. 3. Create a graph by using the m-edges of the input graph of "Starting Node, Terminating Node, Weight Value". 4. Output BFS traversal starting from node 1. In the traversal process, if there are multiple nodes to choose, the node with smaller number will be preferred. 5. Output DFS traversal starting from node 1. In the traversal process, if there are multiple nodes to choose, the node with smaller number will be preferred. 6. Output the length of the shortest path from the first node to the n node. If there is no path, output 0.)