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POJ 1001 F AC source code, do not believe you can try.
ＰＯＪ １００１己ＡＣ源代码，不信你可以试试．-POJ 1001 F AC source code, do not believe you can try.
- 2022-02-07 03:06:09下载
- 积分：1
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poj1125 binary heap using the Dijkstra algorithm optimization
poj1125用二叉堆优化的Dijkstra算法-poj1125 binary heap using the Dijkstra algorithm optimization
- 2022-07-15 13:25:37下载
- 积分：1
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数据结构的最短路径算法
数据结构的最短路径算法-Source codes about how to realize the short path generation for single source graph.
- 2022-06-20 23:37:38下载
- 积分：1
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病人管理系统
病人管理系统是用来保持病人预约和病史。
- 2022-02-15 00:20:35下载
- 积分：1
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one of Huffman coding procedures, using the c series. For a few from the right c...
一个关于哈夫曼编码的程序，用c++编的。可用于对数距的编码-one of Huffman coding procedures, using the c series. For a few from the right coding
- 2023-04-21 14:45:02下载
- 积分：1
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Based on Visual C++ Developed in the course of its work on the elevator simulati...
基于Visual C++开发的，关于电梯工作过程的模拟的程序源代码。电梯模拟-Based on Visual C++ Developed in the course of its work on the elevator simulation program source code. Elevator Simulation
- 2022-03-06 16:08:13下载
- 积分：1
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springdata
这是一个spingdata的项目示例。详细介绍了springdata的配置，注解等。项目完成了对数据库的曾删改查等操作，想要学习springdata的可疑下载参考。
- 2023-07-11 18:05:06下载
- 积分：1
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(1)输入E条弧<j,k>,建立AOE
(1)输入E条弧,建立AOE-网的存储结构 (2)从源点v出发,令ve[0]=0,按拓扑排序求其余各项顶点的最早发生时间ve[i](1=2). (4)根据各顶点的ve和vl值,求每条弧s的最早发生时间e(s)和最迟开始时间l(s).若某条弧满足条件e(s)=l(s),则为关键活动.-(1) E importation of Arc lt; J, kgt; Establish AOE- network storage structure (2) v starting point source, ve [0] = 0, by topological sorting point for the rest of the earliest timing ve [i] (1LT ; = ilt; = n-1). if the Topography vertex orderly sequence number is less than net n vertices, a statement that net presence in Central, not for Critical Path, algorithm implementation steps to terminate or (3) (3) from the Department of v starting point, Vl [n-1] = ve [n-1], by the inverse order for the remaining topology of the latest occurrence of peak time Vl [i] (n-inversion; = IGT; = 2). (4) According to the apex of ve and Vl value for each s arc of the earliest timing e (s) and the latest starting time
- 2022-03-20 01:39:05下载
- 积分：1
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银行排队系统仿真
假设银行只有一个窗口（排一个队），客户到达时间间隔服从均值为200秒的指数分布，客户在窗口接受服务
的时间服从均值为300秒的指数分布。设计排队模拟系统，在进行了100人服务后，输出平均到达间隔、平均
等待时间、平均服务时间以及平均停留时间。
- 2022-01-24 19:08:32下载
- 积分：1
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使用两个向量的叉积判断两线分割…
使用两个向量的叉积来判断两条线段是否相交-The use of the cross product of two vectors to determine whether two line segments intersect
- 2022-03-24 16:16:40下载
- 积分：1
