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MATLAB
遗传算法计算多目标规划问题
使用简单，语言明了(Multi-objective genetic algorithm is planning to use simple, clear language)
- 2010-09-04 16:34:57下载
- 积分：1
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GA_SVM
利用matlab的gui设计的带有输入和输出功能的界面图形，实现等距线可视性很强，自己独立编写的数据和图形，学习非常值得,实现 ga pca bp和svm压缩图像，matlab语言(Using GUI of MATLAB to design interface graphics with input and output functions, realizing the visibility of isometric line is very strong, and it is very worthwhile to study the data and graphics independently written by oneself, realizing GA PCA BP and SVM compressed image, MATLAB language)
- 2020-06-19 18:40:02下载
- 积分：1
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zcgzxl
最长公共子序列的 c代码，可以直接运行获得结果。(C code for the longest common subsequence can be run directly.)
- 2013-12-04 16:13:33下载
- 积分：1
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observe
视频采集 matlab仿真 这个程序可以参考参考 个人觉得不错(Video capture)
- 2012-05-11 08:21:16下载
- 积分：1
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clusterdigit
数字聚类识别,基于图像模式识别-VC++技术实现书的基础上完成的(Number of cluster identification, image-based pattern recognition technology-VC++ based on the completion of the book)
- 2010-03-02 15:03:35下载
- 积分：1
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GSA_matlab
GSA,万有引力算法用于求解智能优化问题,效果特好!!(GSA, Gravitation intelligent optimization algorithm is used to solve problems, good special effects!!)
- 2009-09-17 20:27:10下载
- 积分：1
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xian
运用matlab软件，编程运算，进行回归分析(Use matlab software, programming operations, the regression analysis)
- 2013-09-14 10:15:19下载
- 积分：1
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Untitled
这个代码是用Matlab写的心形代码，是通过点的形式编写的，看起来很有意思。(This code is written using Matlab heart-shaped code, through the form of written, looks very interesting.)
- 2013-08-29 09:21:33下载
- 积分：1
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3zuoye
某市有一码头，每次仅容一辆船停泊装卸货，由于经常有船等候进港，部分人提出要扩建码头。经过调查历史资料发现，码头平均每月停船24艘，每艘船的停泊时间为24±20小时，相邻两艘船的到达时间间隔为30±15小时，如果一艘船因有船在港而等候1小时，其消耗成本为1000元。经预算，扩建码头大约需要1350万元，故市长决策如下：如果未来五年内停泊船只因等候的成本消耗总和超过扩建码头花费则扩建码头，否则，不予扩建。因此，希望你能够帮助市长做出决策。此问题已知到达的大概时间和大概停泊时间，对于此问题用概率统计的方法来做比较复杂，可用程序随机产生到达时间和停泊时间来模拟未来五年内船的停泊，多次模拟预测停泊情况，以做出决策(。-Problem in a city with a terminal expansion dock, each only allow a boat moored loading and unloading, as often there are waiting boat into port, some people proposed to expand pier. Terminal stopping 24 per month, per vessel berthing time of 24 ± 20 hours, the arrival of two adjacent vessels interval 30 ± 15 hours, if a ship due to boat in the harbor while waiting for one hour, and its consumption cost of 1,000 yuan. The budget of about 13.5 million yuan expansion dock, so the mayor decision as follows: If a ship is moored next five years, the cost of waiting more than the sum of consumption spending is expanding marina pier extension, otherwise, not expanded.)
- 2015-04-17 18:59:46下载
- 积分：1
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Ex11
这个是霍夫曼编码的问题，要求输入只含有ABCD的四句话，且每句8个字符的长度，然后求出四个字符每个的概率，然后当成权值计算霍夫曼编码，我的这个程序已经实现了输入、概率统计、计算出霍夫曼编码存在了bian[]这个数组中(hoffuman decoding)
- 2014-11-20 15:50:36下载
- 积分：1
