nautilus geometry

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)

contains a integerTindicating the total number of test cases. Each test case begins with an integerN, denoting the number of stars in the sky. FollowingNLines, each contains2Integersxi,yi , describe the coordinates ofNStars.1≤T≤ 3≤n≤ −10000≤xi,yi≤10000 All coordinates is distinct. Outputfor Each test case, please output "' YES '" If the Stars can form a regular polygon. Otherwise, Output "' NO '" (both without quotes).Sample Input330 01 11 040 00 11 01 150 00 10 22 22 0Sample Ou

Lining up Time Limit: 2000MS Memory Limit: 32768K Total Submissions: 24786 Accepted: 7767 Description"How am I ever going to solve this problem?" said the pilot.Indeed, the pilot is not facing a easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly through the area once in a straight line, and she had to fly over as many points as Pos Sible. All points were given

; Else if(b[mid]==x) pos=mid,r=mid-1; Elser=mid-1; } returnpos+1;}BOOLCMP (X a,x b) {returna.lB.L;}intMain () {scanf ("%d%d",n,k); f[0]=1; for(intI=1; i400000; i++) f[i]=i*f[i-1]%MoD; for(inti=k;i400000; i++) {LL FZ=f[i]%mod,fm=f[k]*f[i-k]%MoD; LL ni=Mod_reverse (FM,MOD); A[i]=fz*ni%MoD; } for(intI=1; i) {scanf ("%d%d", s[i]. l,S[i]. R); B[sz++]=s[i]. L, b[sz++]=S[i]. R B[sz++]=s[i]. L-1; B[sz++]=s[i]. L +1; B[sz++]=s[i]. R1; B[sz++]=s[i]. r+1; } sort (B,b+sz); Sort (S+1, s

Chapter Two, the local theory of the curve2.1 Concept of the curveA discussion of non-regular curves:, which is an irregular point (a sharp point), and it is an irregular curve.Intuitively, discontinuities, outliers, nodes (intersections), cusp points are non-regular points.It is recorded that when the same curve is represented by different parametric equations, the same curve may appear as a regular curve under one parameter representation and a non-regular curve under another parameter represe

; attypedefstructPoint { - intx, y; - Point () {} -Point (intXxintyy): X (xx), Y (yy) {} - }point; - intN, ans; in Point P[MAXN]; - to BOOLCMP (Point A, point B) { + if(a.x = = b.x)returnA.y b.y; - returnA.x b.x; the } * $ BOOLBsintXinty) {Panax Notoginseng intLL =0, rr =n, mm; - while(LL RR) { theMM = (ll + RR) >>1; + if(p[mm].x = = x p[mm].y = = y)return 1; A Else if(CMP (P[MM], point (x, y))) ll = mm +1; the Elserr = mm-1; + - } $ retur

); if(x1==-1x2==-1y1==-1y2==-1) Break; Rec[t].x1= X1,rec[t].y1 = Y1,rec[t].x2=x2,rec[t++].y2 =Y2; X[K]= x1,y[k++] =Y1; X[K]= x2,y[k++] =Y2; } sort (X,x+k); Sort (Y,y+k);}voidsolve () {intT1,t2,t3,t4; for(intI=0; i) {T1=Binary1 (REC[I].X1); T2=Binary1 (REC[I].X2); T3=Binary2 (rec[i].y1); T4=Binary2 (rec[i].y2); for(intj=t1;j){ for(intL = t3;l) {Vis[j][l]=1; } } } intArea =0; for(intI=0; i){ for(intj=0; j) { area+=vis[i][j]* (x[i+1]-x[i]) * (y[j+1]-Y[j]); }} pri

Topic PortalTest instructions: Given some points, ask to make up the area and maximum of two disjoint rectanglesAnalysis: Violent enumeration, first find the two points that can make up the rectangle and save it (vis array very good), and then write a function to determine whether four points inside another rectangle. There was no save rectangle, with for to find the rectangle, the result is confused to forget to judge the situation of the back shape .../*****************************************

Title: The topic is very simple, that is, a matrix is solid, give a line segment, ask if line and matrix intersectProblem-solving ideas: The use of line segments and segments are crossed, and then determine whether the line is inside the matrix, it should be noted that the coordinates of the matrix he gives is obviously not the upper left and right coordinates, it is necessary to judge the bottom left and right point of the coordinates.#include POJ 1410 intersection (computational

;>1; if(Multi (A,LINE[MID].A,LINE[MID].B) >0) l=mid+1; ElseR=mid; } if(Multi (A,LINE[L].A,LINE[L].B) 0) Cnt[l]++; Elsecnt[l+1]++;}intMain () {intN,m,x1,y1,x2,y2; inti,t1,t2; Point A; while(cin>>nN) {cin>>m>>x1>>y1>>x2>>Y2; for(intI=0; i) {cin>>t1>>T2; Line[i].a.x=T1; LINE[I].A.Y=Y1; line[i].b.x=T2; Line[i].b.y=Y2; } memset (CNT,0,sizeof(CNT)); for(intI=0; i) {cin>>a.x>>a.y; S (a,n); } for(intI=0; i) cout": "Endl; coutEndl; } return 0;}POJ 2318 TOYS (computational

first. 1610: [Usaco2008 feb]line tethered game time limit:5 Sec Memory limit:64 MBsubmit:1810 solved:815[Submit] [Status] [Discuss] DescriptionFarmer John has recently invented a game to test the pretentious Bessie. At the beginning of the game, FJ would give Bessie a piece of wood with a non-coincident point of N (2 Input* Line 1th: Enter 1 positive integers: N* 2nd. N+1 Line: Line i+1 with 2 spaces separated by the integer x_i, y_i, describes the coordinates of point IOutputL