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pipe
某石油公司计划建造一条由东向西的主输油管道。该管道要穿过一个有n 口油井的油田。从每口油井都要有一条输油管道沿最短路经(或南或北)与主管道相连。如果给定n 口油井的位置,即它们的x 坐标（东西向）和y 坐标（南北向）,应如何确定主管道的最优位置, 即使各油井到主管道之间的输油管道长度总和最小的位置?证明可在线性时间内确定主管道的最优位置。(An oil company plans to construct a east to west from the main pipeline. The pipeline to pass through an oil field n wells. Wells from each must have a pipeline along the shortest path (or South or North) connected with the competent Road. If given the location of oil wells in n, that is, their x coordinates (east-west) and y coordinates (north-south), should be how to determine the optimal position in charge of Road, even if the wells to the pipeline in charge of Road between the sum of the smallest length location? proved to be linear time to determine the optimal position in charge of Road.)
- 2008-07-07 10:25:25下载
- 积分：1
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Programming-Optimization
Programming Optimization: by Paul Hsieh
- 2012-10-25 20:57:48下载
- 积分：1
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1960_83
是日本船舶教授对船舶附加质量和附加惯性矩的分析。(Is the ship additional mass moment of inertia and additional analysis Japanese ship professor.)
- 2015-03-23 11:20:10下载
- 积分：1
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face-recognition-matlab-program
人脸识别MATLAB程序，识别准确率达到88 。。(Recognition program, face recognition revision, the recognition rate of 88 .)
- 2014-09-11 11:20:57下载
- 积分：1
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PE_Caculation
matlab计算排列熵的程序,可以定量分析混沌或噪声序列的复杂度(a program that is capable of computing the permatation entropy of a time series noise or chaos waveform.)
- 2017-05-03 07:22:56下载
- 积分：1
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VCandMatlab
VC++ and Matlab混合编程的快速实现(VC++ And Matlab Mixed Programming rapid realization of)
- 2009-01-08 14:38:33下载
- 积分：1
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UAV-Cooperative-Decision-and-Control
无人机协同决策与控制英文原版，可供研究无人机协同的同学参考(the English original of UAV Cooperative Decision and Control，for the study of reference UAV cooperative students)
- 2014-05-03 23:06:18下载
- 积分：1
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matcom
说明： 介绍了vc与matlab的联合编程，很有用处(Matlab introduced vc with joint programming, very useful)
- 2008-11-26 16:55:45下载
- 积分：1
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LMS
基于LMS算法的均衡技术在matlab上的仿真应用(Equalization technique based on the LMS algorithm in matlab simulation applications)
- 2013-08-28 21:57:31下载
- 积分：1
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example4_6
：滤波
fs=400 采样频率,必须大于两倍基带信号最高频率
ts=1/fs 采样时间间隔
T=5 时间窗大小
t=0:ts:T (hf1=zeros(1,N)
hf1(f<=B1&f>=-B1)=1
y1=hf1.*fftshift(y)
figure
subplot(211)
plot(f,abs(fftshift(y))/N,f,hf1, r-- )
axis([-fs/2,fs/2,0,1.2])
title( 原信号频谱 )
xlabel( f /Hz )
subplot(212)
plot(f,abs(y1)/N)
title( 低通滤波，截止频率90Hz )
xlabel( f /Hz ))
- 2010-05-25 20:58:05下载
- 积分：1
