octagon geometry

Topic PortalTest instructions: Tell a few rectangles about the area they occupy in a convex polygonAnalysis: Training guide P272, rectangular area long * wide, as long as the calculation of all points, with convex hull and then to find the polygon area. At the center of the known rectangle, the vector rotates at the origin reference point and the angle is converted to radians./************************************************* author:running_time* Created time:2015/11/10 Tuesday 10:34:43* Fi Le N

Round and rectangular area intersectionor triangulation to do it.Code:#include ZOJ 2675 Little Mammoth (computational geometry)

] DescriptionThere are n linear l1,l2 on the Xoy Cartesian plane,... Ln, if the Y value is positive infinity down, you can see a sub-segment of Li, it is said that Li is visible, otherwise li is covered.For example, for a line:L1:y=x; L2:y=-x; L3:y=0Then L1 and L2 are visible, and L3 are covered.gives a line of N, expressed as a form of y=ax+b (| a|,| b|InputThe first behavior n (0 OutputFrom small to large output visible line number, 22 is separated by a space, the last number must also hav

http://poj.org/problem?id=2398This problem and the previous toys is the same is the output is not the same as this gives is the chaos you have to sort the bezel first#include #include#include#include#include#include#include#include#includeusing namespacestd;#defineMemset (A, B) memset (A,b,sizeof (a))#defineN 5500typedefLong Longll;structnode{intx, Y, V;} P[n],u[n];intY2;intcmpConst void*a,Const void*b) { structNode *c, *F; C=(structNode *) A; F=(structNode *) b; returnC->x-f->x;}intFind (int

I believe that in the face of procedural problems involving geometric operations, there is no need to draft a direct strike out the correct code of the great God is present, of course I am not.Recently encountered a practical problem, need to use the knowledge of trigonometric functions learned in junior high school to solve, the problem is described by trigonometric function is in a right triangle, known as a angle α angle, and the angle α of the adjacent edge length A, the angle α of the side

, point + points +1, p1)-point;intEnd = Lower_bound (Point +1, point + points +1, p2)-point-1; for(inti = start; I if(InRange (POINT[I].Y)) ++ans[point[i].id]; }}square[max];structcircle{point P;DoubleRvoidRead () {p.read ();scanf("%LF", r); }BOOLInRange (ConstPoint a) {returnCalc (a,p) voidCount () {intStart = Upper_bound (Point +1, point + points +1, Point (P.x-r,0))-point;intEnd = Lower_bound (Point +1, point + points +1, point (p.x + R,0))-Point-1; for(inti = start; I if(InRange (Point[i]))

to 10000. It is guaranteed that the cake and the ant's home don ' t overlap or contact, and the ant's starting point also are not insid E The cake or his home, and doesn ' t contact with the cake or his home.If the ant touches any part of home, then he's at home.Input ends with a line of 0 0. There may is a blank line between the test cases.Outputfor each test case, print the shortest distance to achieve his goal. Round the result to 2 digits after decimal point.Sample Input1 1-1 1 10-1 1 00 2-

Given two straight lines, determine intersection, overlap, or find the intersectionThe topic of the test templateCode:#include POJ 1269 intersecting Lines (computational geometry)

=deg; theModify (1,1, N, (1+ N) >>1); - } Wu voidAsk (intIintLintRintmid) { - if(Mleft r) { Aboutmx + = t[i].x, my + =t[i].y; $ return; - } - pushdown (i); - if(Mright >= l mleft mid) AAsk (I 1, L, Mid, (L + mid) >>1); + if(Mleft mid) theAsk (I 1|1, Mid +1, R, (R + mid +1) >>1); - } $ voidAskxy (intLeftintRight ) { theMleft = left, mright = right; MX = my =0; theAsk (1,1, N, (1+ N) >>1); the if(ABS (MX) 0.001) MX =0; the ifABS (MY) 0.001) My =0; - } in intMain (

(tem2, tem3); - if(Limit 2) Tem3 = DFS (U +2, U +1, Limit,0); AboutTEM2 =Max (tem2, tem3); $ } - returnMax (tem1, tem2); - } - A intMain () { + //freopen ("In.txt", "R", stdin); the intT; -scanf"%d", T); $ while(t--){ thescanf"%d", n); the inti; theRep (I,0, N) scanf ("%LF", a[i]); theSort (A, A +n); -A[n] =0; inMini = (Double) Linf; theRep (I,1, n) Mini = min (Mini, A[i]-a[i-1]); theMaxi = (Double) Linf; AboutRep (I,1N1) maxi = min (Maxi, max (A[i +1]-a[i

Topic PortalTest instructions: Two dogs running on a polyline, the speed is unknown, at the same time to go, at the same time. The difference between the maximum distance and the minimum distance between two dogs during a runAnalysis: Training Guide P261, consider the relative motion, set a stationary, B relative a motion, the relative motion vector: vb-va (can be understood as the speed vector), then is the PA to the line segment Pb-pb+vb-va distance maximum/************************************

Two-dimensional geometric template--learn from Rujia LiuConst double EPS = 1e-10;struct Point {//points defined by double x, y; Point (Double x=0, double y=0): X (x), Y (y) {}};typedef point Vector; The definition of a vector double polar_angle (vector A) {//Vector polar return atan2 (A.Y, a.x);} Double dot (vector A, vector B) {//vector dot product return a.x * b.x + a.y * B.Y;} Double Cross (vector A, vector B) {//vector fork product return a.x * B.Y-A.Y * b.x;} int dcmp (double x) {/

In Cocos2d v3.x, it is not possible to directly invoke the CCDRAWXXX function as in v2.x to draw the geometry.We can use Ccdrawnode or ccrenderer to draw graphics.But the official API manual makes it clear that Ccdrawnode is only recommended for debugging in the game, because changing its state must be removed and then re-added, resulting in poor performance when drawing large numbers of complex geometries.In cocos2d v2.x, direct calls to OpenGL are generally placed in the draw method, but the a

Approximate test instructions:1E3 segments, drawn on a piece of paper, to find out how many segments, (two segments overlap, or end-to-end will be treated as a line)Ideas:Two segments on the same line: they have the same slope, and they are equal to the projection points on the y-axis or x-axis. And then according to the two order of the sequence can be made out.This card accuracy, to use the EPS#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include Copyright NOTICE: This article for

Test instructions: A circle of land, select n points in the circumference, and then 22 lines, ask how many pieces of this piece of land into?Analysis: This problem uses Euler formula, in the plan, v-e+f=2, where V is the number of vertices, E is the number of edges, F is the number of polygons. For this problem, just calculate V and e as well.We enumerate the diagonal from a vertex, the line to the left of the I point, then the right side of the n-i-2 point, then the two sides of the lines on th

$ (-y_1,x_1) $. Three-point co-line The angle is $0$ and $a\;\times\;b=0$. Whether the point is on the ray The angle is $0$ and the point multiplication $\geq\;0$. $ Polygon Area Set the polygon vertices to $p_1,p_2,..., p_n$ in turn. $\large{s=|\frac{\sum_{i=1}^{n-1}\overrightarrow{op_i}\times\overrightarrow{op_{i+1}}+\overrightarrow{op_n}\ times\overrightarrow{op_{1}}}{2}}|$. Point to line perpendicular Point to line perpendicular $d,ed\;\perp\; ab$. Rotate the $\overrightarrow{ab}\pi/2$. Str

Topic Link: 51nod 1264 segment intersectionIf two segments intersect, either one of the endpoints of one segment is met on another segment, or two segments are on a straight line extending across another segment. (If the point P1 is on one side of the line p3p4 and the point P2 is on the other side of the line, it is said that P1p2 crosses the line p3p4.) ）Cross-multiplication can be used to determine whether the P3P1, p3p2 are in different directions in the P3P4 (CIS, counterclockwise) (line se

Title AddressBrief test Instructions:Given the coordinates of two points, and a, B, C of some general linear equation ax+b+c=0, these straight lines act as streets, seeking from one point to another the number of streets to cross. (Not in the street at two o ' All)Thinking Analysis:The street from one point to the other must span the equivalent of 2.1 points on one side of the line and the other on the other side. Only need to take two point coordinates in, a positive one minus. These lines are

Tenth chapter, very little surface1. Minimum figureAnimation showing the deformation of a helicoid into a catenoid.Animation of Scherk ' s first and second surface transforming into each other:they is members of the same associate family of minimal surfaces.1.1. isothermal coordinates of minimum graphs1.2.Bernstein theorem2. Weierstrass expression of a very little surfaceEnneper surface,Costa surface, a complete minimum embedded surface with a genus of 1.3. Gauss mapping of minimal surfacesMerom

Title Address: http://acm.hdu.edu.cn/showproblem.php?pid=3365Idea: Take a[0] as the origin, construct vector a[i]-a[0]. First rotate (note the direction of rotation), then stretch, and finally pan to the end.#include Hdu 3365 New Ground (computational geometry)