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Stack

于 2012-11-16 发布 文件大小:811KB
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  este es el codigo de ejersicio de stack en c++ sin errores esta en zip

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  • xiande
    弗洛伊德算法计算有向网络图最短路问题。可算最短路径。(Freud algorithm to calculate the shortest path problem to the network diagram. Can be considered the shortest path.)
    2012-08-17 14:02:44下载
    积分:1
  • new-DV-Hop-code
    改进DV-Hop定位算法 首先设置初始量,布置了一个范围为100×100m2的区域,其上随机分布100个传感器节点,其中有10个信标节点,节点的通信半径为30m。 第二步在正方形区域内产生均匀分布的随机拓扑,随机产生节点坐标并将其中十个选定为信标节点,其余九十个设为未知节点,然后画出节点分布图。 第三步通过最短路径法计算未知节点与每个信标节点的最小跳数。 第四步根据前面记录的其他信标节点的位置信息和相距跳数估算平均每跳的实际距离,用跳数估计距离的方法得出未知节点到信标节点的距离。 第五步用极大似然估计法求未知节点坐标 (Improved DV-Hop localization algorithm first set the initial amount, layout 100100m2 the area of ​ ​ a range of 100 sensor nodes randomly distributed on the 10 beacon node, the node communication radius of 30m. The second step in the square area to generate uniformly distributed random topology, random coordinates of the nodes and ten of the selected beacon node, the remaining 90 is set to unknown node, and then draw the node distribution diagram. The third step is to calculate the minimum number of hops of the unknown node and each beacon node through the shortest path method. The fourth step according to the location information of the other beacon nodes in the previous record and away from hops to estimate the average hop distance and hop count to estimate the distance to come to the distance of the unknown node to beacon nodes. The fifth step maximum likelihood estimation method and the unknown coordinates of the nodes)
    2020-10-12 22:27:31下载
    积分:1
  • very handy calculator that can calculate a lot of things, including some simple...
    非常好用的计算器,可以计算很多东西,包括一些简单函数的求解-very handy calculator that can calculate a lot of things, including some simple function of solution
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  • 一个快速排序算法的实现例子,课程作业,可作研究学习之用。 包含源文件和详细的开发文档,可方便移植到C等语言平台。...
    一个快速排序算法的实现例子,数据结构课程作业,可作研究学习之用。 包含源文件和详细的开发文档,可方便移植到C等语言平台。-a fast algorithm to achieve example, the data structure course work and can be used for research and study purposes. Includes source document and detailed documentation, will facilitate the transplantation of languages such as C platform.
    2022-04-15 16:19:00下载
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    2022-03-14 02:54:06下载
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    问题描述:平均路径长度是网络中另一个重要的特征度量,他是网络中所有节点对之间的平均最短距离。这里节点间的距离指的是从一个节点要经历的边的最小数目,其中所有节点之间的最大距离称为网络的直径。平均路径长度和直径衡量的是网络的传输性能与效率。平均路径长度的公式为 ,其中dij表示点i和j之间的最短距离(若dij不存在时,dij就不能加入,且分母要相应减1) 要求:输入邻接矩阵表示的图,计算其平均路径长度(Description of the problem: the average path length is another important network characteristics measure between all pairs of nodes in the network, the average shortest distance. Here the distance between nodes refers to the minimum number from one node to be subjected to the edge, wherein the maximum distance is referred to as the diameter of the network between all nodes. The average path length and diameter measurement of the performance and efficiency of the network' s transmission. The formula of the average path length, where dij indicates the shortest distance between points i and j (if dij is not present, dij can not join, and the denominator accordingly minus 1) requirements: Input adjacency matrix shows, to calculate its average path length)
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