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Description: 最接近点对问题是求二维坐标中的点对问题，该算法是为了将平面上点集S线性分割为大小大致相等的2个子集S1和S2，我们选取一垂直线l:x=m来作为分割直线。其中m为S中各点x坐标的中位数。由此将S分割为S1={p∈S|px≤m}和S2={p∈S|px>m}。从而使S1和S2分别位于直线l的左侧和右侧，且S=S1∪S2 。由于m是S中各点x坐标值的中位数，因此S1和S2中的点数大致相等。
递归地在S1和S2上解最接近点对问题，我们分别得到S1和S2中的最小距离δ1和δ2。现设δ=min(δ1,δ1)。若S的最接近点对(p,q)之间的距离d(p,q)<δ则p和q必分属于S1和S2。不妨设p∈S1，q∈S2。那么p和q距直线l的距离均小于δ。因此，我们若用P1和P2分别表示直线l的左边和右边的宽为δ的2个垂直长条，则p∈S1，q∈S2。
-closest point to the problem is for two-dimensional coordinates of the point of the problem, the algorithm is to be planar point set on the S linear divided roughly equal to the size of two sub-sets of S1 and S2, we selected a vertical line l : x = m as a separate line. Where m for S x coordinates of the points of the median. This will be divided into S S1 = (p S | px m) and (p = S2 S | pxgt; M). So that S1 and S2 are located in the linear l left and right, and S = S1 S2. M is due to S x coordinates of the points of the median value, S1 and S2 2186 roughly equal. Recursive in S1 and S2 the nearest point on the solution of the problem, we won the S1 and S2 the minimum distance between 1 and 2. The existing = min (1, 1). If S is the closest point of (p, q) the distance between d (p, q)
Platform: |
Size: 15666 |
Author: 黄波 |
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Description: 1. 根据csv文件数据内容绘图。csv文件内容包括：图形类型（横线，竖线，横半圆弧，竖半圆弧），长度（线长度，半圆半径）。根据csv文件中提供的数据，在dialog中绘制出相应的图形。如果要绘制的图形超出dialog的 x轴边界，则从x=0，y不变的坐标开始绘制该图形；如果超出了y轴边界，则从x不变，y=0坐标开始绘制。-1. According csv document mapping data elements. Csv file contents include : graphic types (horizontal line, vertical line, horizontal semi-circular, semi-circular vertical), the length (length, semicircle radius). Csv document according to the information provided by the dialog in drawing up the corresponding graphics. If the Graphics Rendering dialog beyond the x-axis border, from x = 0, y unchanged began mapping the coordinates of the graphics; If we look beyond the borders of the y-axis, from x unchanged. y = 0 start mapping coordinates.
Platform: |
Size: 90457 |
Author: 李小跳 |
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Description: 此算法为多传感器目标定位法，是基于实现交叉定位，采用公垂线法。-algorithm for the multi-sensor target identification, is based on the realization of cross-positioning, use common vertical line method.
Platform: |
Size: 2742 |
Author: 李娴 |
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Description: 线段及直线的基本运算 1. 点与线段的关系2. 求点到线段所在直线垂线的垂足3. 点到线段的最近点4. 点到线段所在直线的距离 5. 点到折线集的最近距离 6. 判断圆是否在多边形内7. 求矢量夹角余弦 8. 求线段之间的夹角9. 判断线段是否相交10.判断线段是否相交但不交在端点处11.求线段所在直线的方程 12.求直线的斜率 13.求直线的倾斜角14.求点关于某直线的对称点15.判断两条直线是否相交及求直线交点 16.判断线段是否相交，如果相交返回交点-straight line segment and a basic computing. Point and the line between the two. Seeking to point straight vertical line segment where the Pedal 3. Line points to the recent four points. Point to the straight line distance from the host 5. Set point to the recently broken away from six. To judge whether the circular polygon 7. for vector cosine angle 8. for the angle between the line 9. Line judge whether the intersection 10. Line judge whether the intersection but do not pay the endpoint Office 11. for the straight line segment where equation 12. for a linear slope 13. seeking straight tilt angle 14. Points on a linear symmetrical points 15. to judge whether the intersection of two straight and for 16 straight intersection. Line judge whether the intersection, the intersection to intersec
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Size: 1265 |
Author: 孤星赶月 |
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Description: 线段及直线的基本运算 1. 点与线段的关系2. 求点到线段所在直线垂线的垂足3. 点到线段的最近点4. 点到线段所在直线的距离 5. 点到折线集的最近距离 6. 判断圆是否在多边形内7. 求矢量夹角余弦 8. 求线段之间的夹角9. 判断线段是否相交10.判断线段是否相交但不交在端点处11.求线段所在直线的方程 12.求直线的斜率 13.求直线的倾斜角14.求点关于某直线的对称点15.判断两条直线是否相交及求直线交点 16.判断线段是否相交，如果相交返回交点-straight line segment and a basic computing. Point and the line between the two. Seeking to point straight vertical line segment where the Pedal 3. Line points to the recent four points. Point to the straight line distance from the host 5. Set point to the recently broken away from six. To judge whether the circular polygon 7. for vector cosine angle 8. for the angle between the line 9. Line judge whether the intersection 10. Line judge whether the intersection but do not pay the endpoint Office 11. for the straight line segment where equation 12. for a linear slope 13. seeking straight tilt angle 14. Points on a linear symmetrical points 15. to judge whether the intersection of two straight and for 16 straight intersection. Line judge whether the intersection, the intersection to intersec
Platform: |
Size: 1024 |
Author: 孤星赶月 |
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Description: 最接近点对问题是求二维坐标中的点对问题，该算法是为了将平面上点集S线性分割为大小大致相等的2个子集S1和S2，我们选取一垂直线l:x=m来作为分割直线。其中m为S中各点x坐标的中位数。由此将S分割为S1={p∈S|px≤m}和S2={p∈S|px>m}。从而使S1和S2分别位于直线l的左侧和右侧，且S=S1∪S2 。由于m是S中各点x坐标值的中位数，因此S1和S2中的点数大致相等。
递归地在S1和S2上解最接近点对问题，我们分别得到S1和S2中的最小距离δ1和δ2。现设δ=min(δ1,δ1)。若S的最接近点对(p,q)之间的距离d(p,q)<δ则p和q必分属于S1和S2。不妨设p∈S1，q∈S2。那么p和q距直线l的距离均小于δ。因此，我们若用P1和P2分别表示直线l的左边和右边的宽为δ的2个垂直长条，则p∈S1，q∈S2。
-closest point to the problem is for two-dimensional coordinates of the point of the problem, the algorithm is to be planar point set on the S linear divided roughly equal to the size of two sub-sets of S1 and S2, we selected a vertical line l : x = m as a separate line. Where m for S x coordinates of the points of the median. This will be divided into S S1 = (p S | px m) and (p = S2 S | pxgt; M). So that S1 and S2 are located in the linear l left and right, and S = S1 S2. M is due to S x coordinates of the points of the median value, S1 and S2 2186 roughly equal. Recursive in S1 and S2 the nearest point on the solution of the problem, we won the S1 and S2 the minimum distance between 1 and 2. The existing = min (1, 1). If S is the closest point of (p, q) the distance between d (p, q)
Platform: |
Size: 15360 |
Author: 黄波 |
Hits:
Description: 1. 根据csv文件数据内容绘图。csv文件内容包括：图形类型（横线，竖线，横半圆弧，竖半圆弧），长度（线长度，半圆半径）。根据csv文件中提供的数据，在dialog中绘制出相应的图形。如果要绘制的图形超出dialog的 x轴边界，则从x=0，y不变的坐标开始绘制该图形；如果超出了y轴边界，则从x不变，y=0坐标开始绘制。-1. According csv document mapping data elements. Csv file contents include : graphic types (horizontal line, vertical line, horizontal semi-circular, semi-circular vertical), the length (length, semicircle radius). Csv document according to the information provided by the dialog in drawing up the corresponding graphics. If the Graphics Rendering dialog beyond the x-axis border, from x = 0, y unchanged began mapping the coordinates of the graphics; If we look beyond the borders of the y-axis, from x unchanged. y = 0 start mapping coordinates.
Platform: |
Size: 4040704 |
Author: 李小跳 |
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Description: 此算法为多传感器目标定位法，是基于实现交叉定位，采用公垂线法。-algorithm for the multi-sensor target identification, is based on the realization of cross-positioning, use common vertical line method.
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Size: 2048 |
Author: 李娴 |
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Description: 分治法解决最近对问题
画一条垂直线x=c，把这些给定点分为两个包含n/2个点的子集S1和S2，使得n/2个点位于直线的左侧或直线上，另外n/2个点位于直线的右侧或直线上；遵循分治法的思想，递归地求出左子集S1和右子集S2中的最近对，分别为d1与d2；之后d=min{d1,d2}。合并过程：在以垂线x=c为对称轴，2d为宽度的区域内求最近两个点的距离，记为d3；求D=min{d,d3}；
-Sub-rule method to resolve the issue of the recent draw a vertical line x = c, a given point of these two contains n/2 points in a subset of S1 and S2, making n/2 points located on the left side of a straight line or straight line the other n/2 points located on the right side of a straight line or straight line follow the idea of sub-rule method, recursive subset to derive the left and right S1 subset S2 of the recent right, respectively, d1 and d2 after d = min (d1, d2). The merger process: in the vertical line x = c as the axis of symmetry, 2d is the width of the region for the last two points the distance is recorded as d3 for D = min (d, d3)
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Size: 1024 |
Author: 王鹏 |
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Description: （1）用Bresenham算法绘制直线
（2）程序适用各种直线情况，包括水平线，垂直线
（3）采用键盘输入线段的两个端点的坐标
-(1) using Bresenham algorithm for drawing a straight line (2) procedure applies to all kinds of straight lines, including horizontal, vertical lines (3) the use of keyboard input segment of the two endpoint coordinates
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Size: 1024 |
Author: Eean |
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Description: 用光电耦合器测量出重物下降过程中经过任何等距离且在同一垂直线上的三点时间间隔t1,t2,及距离h，然后利用公式g=2h(t1-t2)/t1t2(t1+t2)使用浮点数运算方法计算出g的值。最后由单片机p3口送串行显示。-Optocoupler used to measure the process of heavy decline any equidistant and at the same vertical line of three time intervals t1, t2, and the distance h, and then use the formula g = 2h (t1-t2)/t1t2 (t12B ! t2) the use of floating-point method to calculate the value of g. Finally, I sent Singlechip p3 serial display.
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Size: 5120 |
Author: 张品 |
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Description: 在图形模式下显示鼠标，以鼠标坐标位置为基准点在屏幕上画一条纵贯屏幕的垂直线和一条横贯屏幕的水平直线。当鼠标移动时，两条直线也跟着移动。当按下鼠标右键时停止程序运行-In graphics mode mouse to the mouse coordinates of the location for the reference point on the screen draw a vertical line runs through the screen and a straight line running through the level of the screen. When the mouse moves, the two also followed a straight line moving. When pressing the right mouse button to stop running when
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Size: 1024 |
Author: lg |
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Description: 一线总线协议可以在一条纵线上搜索多个DS18B20的ROM-First-line bus protocol can be in a vertical line of the ROM search multiple DS18B20
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Size: 2048 |
Author: maomao |
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Description: 该程序为一arm参考程序,实现在液晶屏上画垂直线,水平线和直线-One arm of the procedure reference procedures realize the LCD screen painting vertical lines, horizontal and straight-line
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Size: 1024 |
Author: 宁晓平 |
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Description: LCD触摸显示屏控制定位系统:LCD触摸屏也变得普及，并逐渐成为当今的主流配置。触摸屏分为电阻、电容、表面声波、红外线扫描和矢量压力传感触摸屏等，其中使用最多的是四线或五线电阻触摸屏。四线电阻触摸屏是由两个透明电阻膜构成的，在它的水平和垂直电阻网上施加电压，就可通过A/D转换面板在触点测量出电压，从而对应出坐标值。-LCD touch screen control positioning system: LCD touch screen has become popular, and has gradually become the mainstream of today s configuration. Touch screen is divided into resistors, capacitors, surface acoustic wave, infrared scanning and vector pressure sensor touch screens, which use the most is the four-or five-wire resistive touch screen. Four-wire resistive touch screen is made up of two transparent membrane consisting of resistance, in its horizontal and vertical line resistance imposed voltage, can be through the A/D conversion panel measured in the contact voltage, which corresponds to the coordinate values.
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Size: 10240 |
Author: 刘方 |
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Description: 计算水平线与垂直线交点个数问题。扫描方法计算。-Calculation of horizontal and vertical line of intersection the number of questions. Scanning method.
Platform: |
Size: 1024 |
Author: 黄半仙 |
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Description: 线段检测器和线段检测方法。包含垂直线检测部分，水平线检测部分-Detector and line segment detection methods. Contains the vertical line detecting part detecting part of the horizon, etc.
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Size: 2936832 |
Author: 小鹿 |
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Description: 本程序通过对图像进行灰度化、去噪和边缘提取，能检测出不垂直的刻度线边缘并能将其旋转校正，从而达到使刻度线标准化的目的-This procedure of gray-scale images, and de-noising and edge detection can not detect the edge of the vertical line of the scale and be able to spin correction, so as to achieve so that the purpose of standardization of calibration line
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Size: 3072 |
Author: sunhong |
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Description: 对一段语音进行端点检测并对检测的语音画出方框图，不同于一般只画竖线-Endpoint detection of a voice to draw the block diagram and detection of voice is different from the vertical line is generally drawn
Platform: |
Size: 1024 |
Author: hu |
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Description: C# 画竖线的控件，运行生成d-Draw a vertical line of control, run the generated dll
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Size: 34816 |
Author: 杜洋 |
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