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Description: 使用SSD（sum squared differences）算法进行双摄像机图像匹配点计算，基于OpenCv的C++代码实现-Using SSD (sum squared differences) algorithm dual camera image matching points calculation, based on the OpenCv the C++ Code
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Size: 1866752 |
Author: 梁霄 |
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Description: 对于空间两个点云，利用经典的icp算法进行匹配-ICP fit points in data to the points in model. Fit with respect to minimize the sum of square errors with the closest model points and data points.
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Size: 5120 |
Author: wangwei |
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Description: 基于多核的并行算法设计，计算一个连续数列的和-The parallel algorithm based on multi-core design, computing a continuous series and
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Size: 2406400 |
Author: tstao |
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Description: 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-A city with a terminal, each time only to allow a loading and unloading cargo vessels, as vessels waiting to enter port often, some people proposed expansion of the dock. After investigation found that historical data, the terminal stop for 24 per month on average, each ship s turnaround time of 24 ± 20 hours, adjacent to two ship s arrival time interval of 30 ± 15 hours, if a ship due to ship in Hong Kong The wait for 1 hour, its consumption costs of 1,000 yuan. The budget, expansion of the dock takes about 13.5 million yuan, so the mayor the decision-making are as follows: If the next five years, anchored ships were waiting for more than the sum of the cost of consumption spending for expansion of the dock expansion of the dock, otherwise, not expansion. Therefore, I hope you can help the mayor make decisions. This problem is known to the approximate arrival time and approximate turnaround time for this issue to do with the probability of more complicated statistical methods can be u
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Size: 1024 |
Author: camel |
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Description: 首先介绍了LDPC码的校验矩阵和其因子表示方法，然后利用二分图对和积解码算法进行了详细的描述，最后给出了信度传播概率译码算法详细步骤，并对关键公式作了证明-This paper,first introduces the check matrix and the factor graph of LDPC,then describes the
sum-product algorithm by using the factor graph,and finally presents the detailed steps of the sum-product
algorithm and gives a proof of certain important expressions.
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Size: 524288 |
Author: 秦小星 |
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Description: clear all
I=imread( lena.bmp )
figure imshow(I)
I2=imrotate(I,-4, bilinear ) 逆时针旋转4度
figure imshow(I2)
I3=fliplr(I) 垂直镜像
figure imshow(I3)
I4=imresize(I,0.5, bilinear ) 缩小为原图的1/2
figure imshow(I4)
A=double(I) 转换为double类型
计算7个不变矩
[nc,nr]=size(A)
[x,y]=meshgrid(1:nr,1:nc) 得到网格
x=x(:)
y=y(:)
A=A(:)
m.m00=sum(A)
if m.m00==0
m.m00=eps
end
m.m10=sum(x.*A)
m.m01=sum(y.*A)
计算均值
xmean=m.m10/m.m00
ymean=m.m01/m.m00
计算中心矩-li9_23.rar
cm.cm00=m.m00
cm.cm02=(sum((y-ymean).^2.*A))/(m.m00^2)
cm.cm03=(sum((y-ymean).^3.*A))/(m.m00^2.5)
cm.cm11=(sum((x-xmean).*(y-ymean).*A))/(m.m00^2)
cm.cm12=(sum((x-xmean).*(y-ymean).^2.*A))/(m.m00^2.5)
cm.cm20=(sum((x-xmean).^2.*A))/(m.m00^2)
cm.cm21=(sum((x-xmean).^2.*(y-ymean).*A))/(m.m00^2.5)
cm.cm30=(sum((x-xmean).^3.*A))/(m.m00^2.5)
im(1)=cm.cm20+cm.cm02
im(2)=(cm.cm20-cm.cm02)^2+4*cm.cm11^2
im(3)=(cm.cm30-3*cm.cm12)^2+(3*cm.cm21-cm.cm03)^2
im(4)=(cm.cm30+cm.cm12)^2+(cm.cm21+cm.cm03)^2
im(5)=(cm.cm30-3*cm.cm12)*(cm.cm30+cm.cm12)...
*((cm.cm30+cm.cm12)^2-3*(cm.cm21+cm.cm03)^2)...
+(3*cm.cm21-cm.cm03)*(cm.cm21+cm.cm03)...
+(3*(cm.cm30+cm.cm12)^2-(cm.cm21+cm.cm03)^2)
im(6)=(cm.cm20-cm.cm02)*((cm.cm30+cm.cm12)^2-(cm.cm21+cm.cm03)^2)...
+4*cm.cm11*(cm.cm30+cm.cm12)*(cm.cm21+cm.cm03)
im(7)=(3*cm.cm21-cm.cm03)*(cm.cm30+cm.cm12)...
*((cm.cm30+cm.cm12)^2-3*(cm.cm21+cm.cm03)^2)...
+(3*cm.cm12-cm.cm30)*(cm.cm21+cm.cm03)...
*(
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Size: 1024 |
Author: lzp_llz |
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