polygon diamonds

1. Note that polygons in polygon elements can only be planar and must be convex, and the edges of the polygons cannot be bent2. Segmentation and boundaries, which can be artificially set for boundary and non-boundary edgesGledgeflag (TRUE)//Next all points are boundary edge beginningsGledgeflag (FALSE)//Next all points are non-boundary edge start3. Manipulating depth buffersGlutinitdispalymode (glut_depth);//Open Depth bufferGlenable (gl_depth_test);/

InputThe input data contains multiple test instances, one row per test instance, and the beginning of each line is an integer n (3All integers in the input data are within a 32-bit integer range, and n=0 represents the end of the data and does not handle it.OutputFor each test instance, output the corresponding polygon area, and the result is exactly one decimal point after the decimal.The output for each instance takes up one row.Sample Input3 0 0 1

DescriptionThere is a convex P-shape (pNot for (0,0), (0,m), (n,0), (n,m). Each time you can select a line to cut the current shape into two parts, preserving one part (anotherPart of the length of the cut line in the inner part of the polygon for this straight line. Find the shortest cutting line total length. Here is an example. WeNeed to get the middle polygon.Cut along a straight line 1,2,3,4, respectively, to get the middle quadrilateral.InputThe

Polygon Center of Gravity Template # Include # Include # Include Using namespace STD; Struct point{Double X;Double Y;}; // Calculate the center of gravity of a polygonAlgorithm// Description:// Finding the center of gravity of a polygon is not simply extending the formula of the center of gravity of a triangle.// My algorithm is to take a point on the plane (generally taking the origin, this can reduc

Question: starting from a point and eight directions, give the direction of each step and find the area of the polygon formed by the path. However, the question says that we will use the vector cross product to solve the Polygon Area at the beginning and end, and it is not just applicable to convex edges. The concave part is offset by positive and negative.However, this question does not hate precision.Doub

Mathematical Solutions:Select a point outside the polygon, connect each point to form a triangle, calculate sum ... Detailed Link http://blog.csdn.net/hemmingway/article/details/7814494Three sides of the known triangular length, triangular area------> Helen Formula Link: http://www.zybang.com/question/29d209e9732d1f1f6a6de35b94edd3ba.htmlImplemented by the following Python code source code sources link: http://blog.csdn.net/zfq43/article/details/85952

how to draw a straight line, then the square, rectangle, polygon is good to do.Let's start with a poly (e.g. Pentagon) chestnut . Use the existing knowledge to think about, how to achieve?It's simple, as long as there are five vertex coordinates, you can connect them with straight segments:Override func DrawRect (rect:cgrect) {//Pentagon let color = Uicolor.redcolor () color.set ()//Set Line color let Apath = Uibezierpath () apath

Improved scanning-line filling algorithm The principle and implementation of the scan line filling algorithm is simple, but the requirement for storage space is high because it maintains both the active side table (AET) and the New Edge table (NET). Part of the two tables is duplicated, and the "New Edge table" is in many cases a sparse table, if you can improve it, avoid two tables, you can save storage space, while eliminating from the "side table" to generate "new side table" overhead, but a

The planar region filling algorithm is an important algorithm in the field of computer graphics, where the region fill gives the boundary of a region (or it can be without boundary, but gives the specified color), and requires that all pixel elements within the boundary range be modified to the specified color (or pattern fill). The most commonly used in area filling is the polygon fill, in this paper we discuss several

For your illustrator software users to detailed analysis to share the design of stylish cool polygon Poster Quick Drawing Tutorial. Tutorial Sharing: Okay, the above information is small make up to you illustrator of this software users to bring the detailed design fashion cool polygon posters of the rapid drawing of the entire content of the tour, the users see here the small

Low Poly is a lower precision model in 3D, which is used in graphic design because it is a low polygon portrait in portrait.Production process: First find a pair of pixels higher image, intercept the desired part. Then in PS according to portrait contour draw triangle and polygon combination, finished into the AI, and then use the pen to draw the vector grid with the hand-drawn graphics. Finally, fill in th

HDU 1115:http://acm.hdu.edu.cn/showproblem.php?pid=1115 To give you a n, there are n points, and then give you the coordinates of n points, to find the n point of the polygon formed by the coordinates of the center of gravity. More Wonderful content: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/ struct point {double x, y;} P[1000010]; struct Line {point A, b;}; Double Xmult (Point P1, point P2, point P) {return (p1.x-p.x)

Original address: http://www.html-js.com/article/1538 (front end stew) Ray method is a very simple and intuitive way to judge whether a plane interior point is inside a polygon. In addition to the Ray method there are many other methods, today we will introduce a way to judge by the number of rotation. A closed curve in a plane is the number of rotation (also called a winding number) of a point, representing the total number of times the curve bypasse

1. Use Matlab as a closed polygon drawing Instead of finding a function that directly draws a polygon, you can only save the coordinates of each point in an array, connect a point to its adjacent point, and attach the last point to the first point. The following is an example of the. m File: Copy Code code as follows: Clear CLC A=[0 2 4 6 8 10 12 14;0 2 1 4 6 6 5 7]; % the point coordi

Freeform Polygon Intersection: Private voidBtnpolygon_click (Objectsender, EventArgs e) { Double[] Xs =New Double[] { -,260, the, -, -, the, the, the, -, -, -, -, -,260 }; Double[] Ys =New Double[] { $,260, the, Max, the, About, $,260 }; intm = xs.length/2; varcoords =NewCoordinate[m]; varCoordscheck =NewCoordinate[m]; for(vari =0; I 2; i++) { varx = Xs[i *2]; vary = xs[i *2+1]; Coords[i]=Newcoordinate (x, y); Coo

1. Create a polygon based on 4 pointsCreate a graph based on 4 points , order clockwise1234public static Ipolygon createpolygonby4points (IPoint pnt1, IPoint pnt2, IPoint pnt3, IPoint pnt4){Ipointcollection Ppntcol = new Polygonclass ();Object missing = Type.Missing; Add ClockwisePpntcol.addpoint (pnt1, ref missing, ref missing);Ppntcol.addpoint (Pnt2, ref missing, ref missing);Ppntcol.addpoint (PNT3, ref missing, ref missing);Ppntcol.addpoint (PNT4,

Test instructions: A bar of length n, randomly select K position to cut it into k+1 section, ask this k+1 section can make up the probability of the polygon k+1 edge . Analysis: This problem, obviously and N is not any relationship, because no matter how much n is how many segments can be, only with the number of segments of the cut. Then we have to convert it, can't do it directly, because it's not good to do.Convert to a circle to the election m+1 p

Using Dotspatial.controls;Using Dotspatial.data;Using DotSpatial.Data.Forms;Using DotSpatial.Plugins.Measure;Using Dotspatial.symbology;Using Dotspatial.topology;Using DotSpatial.Topology.Algorithm;Using point = DotSpatial.Topology.Point;Using Pointshape = DotSpatial.Symbology.PointShape;Shapefiles s = new Shapefile ();Polygon FeaturesFeatureSet fs = new FeatureSet (Featuretype.polygon);Add FieldFs. DATATABLE.COLUMNS.ADD (New DataColumn ("ID", typeof

When creating a polygon div, the younger brother encountered a problem: he wanted to generate a polygon div. the front div protruding a triangle, and the back div was concave into a triangle, embed and overlap the protruding and concave parts of the two divs. how can this problem be achieved ?, It is implemented by image spoofing. PHP self-painting is not beautiful. canvas painting is not compatible. it is

After a simple polygon is generated, it is sometimes necessary to judge the concave and convex of each vertex of the polygon. Calculate the two vectors of the vertex to be processed and the adjacent vertex, and then calculate the cross multiplication of the two vectors. Based on the positive and negative values of the obtained result, we can judge the concave and convex. If the result is negative, it is a c