octagon geometry

The sum of the areas of all triangles that can be composed of the points on a given planeFirst we enumerate each point to establish a planar Cartesian coordinate with this point as the origin and then sort the points on the first to fourth quadrant and the X, Y axis positive half axes according to the slope.Enumerating the second and third points to do this is an O (n^3) positive timeout But what did we find?For each point K its contribution to the answer is:(X1*YK-Y1*XK) + (X2*YK-Y2*XK) +...+ (

This problem with n^2 algorithm can pass, the first arbitrary enumeration of two points, and the center of the triangle to calculate the area, this area may be added (n-2) times, but note that if there are 3 points in the same side, then to subtract, so in the enumeration, each time enumerating a point, and then enumeration and the point of the degree of the difference between 180 points For area, this area minus 2 * (j-i + 1) timesCode:#include UVA 11186-circum Triangle (computational

Enumerate straight lines, calculate slope, sort, statistical answer.#include "Calculate geometry" "slope" bzoj1610 [Usaco2008 feb]line tethered Game

Turn the cornerTime limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total Submission (s): 2059 Accepted Submission (s): 785problem DescriptionMr West bought a new car! So he's travelling around the city.One day he comes to a vertical corner. The street he is currently in have a width x, the street he wants to turn to have a width y. The car has a length l and a width d.Can Mr West go across the corner?InputEvery line have four real numbers, X, Y, L and W.Proceed to the e

coordinates A (x1,y1) and B (X2,y2) is given by anticlockwise.OutputFor each test case, you should output the coordinate of C (X3,Y3), the result should is rounded to 2 decimal places in a Li Ne.Sample Input4-100.00 0.00 0.00 0.000.00 0.00 0.00 100.000.00 0.00 100.00 100.001.00 0.00 1.866 0.50Sample Output( -50.00,86.60) ( -86.60,50.00) ( -36.60,136.60) (1.00,1.00)The problem and algorithm analysis: input a point coordinate, then enter B point coordinate, calculate c point coordinates, accordin

; - } in return 1; - } to + intMain () { - intT; thescanf"%d",T); * while(t--) { $scanf"%d",n);Panax Notoginsengfor (I,0, N) -scanf"%LF%LF%LF%LF",l[i].x1,l[i].y1,l[i].x2,l[i].y2); the if(n==1) {puts ("yes!");Continue; } + BOOLans=0; Afor (I,0, N) { thefor (j,i+1, N) { + if(Judge (l[i].x1,l[i].y1,l[j].x1,l[j].y1)) ans=1; - if(Judge (L[i].x1,l[i].y1,l[j].x2,l[j].y2)) ans=1; $ if(Judge (l[i].x2,l[i].y2,l[j].x1,l[j].y1

Topic Link: UVA 1308-viva confettiEnumerate the two circles, process all arcs, and then determine whether the midpoint of each arc is visible, the circle where the arc is visible, and the circle below the arc is also visible.#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. UVA 1308-viva Confetti (geometry)

;}//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////three-dimensional vector operationBOOL operator==(Point3D p1, Point3D p2) {return(eq (p1.x, p2.x) eq (p1.y, p2.y) EQ (p1.z, p2.z));}BOOL operator(Point3D p1, Point3D p2) {if(NEQ (p1.x, p2.x)) {return(P1.x p2.x); } Else if(NEQ (P1.Y, p2.y)) {return(P1.y p2.y); } Else { return(P1.z p2.z); }}point3doperator+(Point3D p

DoubleA =2.0* (x.c.x-y.c.x), B =2.0* (X.C.Y-Y.C.Y), C = y.c.x*y.c.x-x.c.x*x.c.x + y.c.y*y.c.y-x.c.y*x.c.y+x.r*x.r-y.r*Y.R; + DoubleL1 =Dis_line (x.c,a,b,c); - DoubleCo1 = ACOs (L1/X.R) *2.0; the DoubleS1 = PI * x.r*x.r* (co1/pi/2.0) - (0.5*x.r*x.r*sin (co1)); * DoubleL2 =Dis_line (y.c,a,b,c); $ DoubleCO2 = ACOs (L2/Y.R) *2.0, S2;Panax Notoginseng if((a*x.c.x+b*x.c.y+c) * (a*y.c.x+b*y.c.y+c) 0)//Two center on both sides of the garden intersection -S2 = PI * y.r*y.r* (

input lines represents, lines on the Plane:the line through (x1,y1) and (X2,y2) and the line Throug H (x3,y3) and (X4,y4). The point (x1,y1) is always distinct from (x2,y2). Likewise with (X3,y3) and (X4,y4).OutputThere should be n+2 lines of output. The first line of output should read intersecting LINES output. There'll then being one line of output for each pair of planar lines represented by a line of input, describing how the Lin Es intersect:none, line, or point. If the intersection is a

;}BOOLZeroDoublex) {returnFabs (x) EPS;} Point_3 Pvec (point_3 s1,point_3 s2,point_3 S3) {returnDet (S1-S2), (S2-S3));}intDots_onplane (point_3 a,point_3 b,point_3 c,point_3 d) {returnZero (Dot (Pvec (A, B, c), D-a));}intN; Point_3 Point3[max_n];intMain () {scanf ("%d", N); for(inti =0; I ) { intA, B, C; scanf ("%d%d%d", a, b, c); Point3[i]=Point_3 (A, B, c); } if(N 3) {cout"TAK"Endl; return 0; } intp =-1; for(inti =2; I ) { if(Dot_inline (point3[0], point3[1], point3[i]) =

[ -];structseg{intL,r,len;} seg[ -];//Data EndsintMain () {intn,i,j; DoubleGen2 = sqrt (2.0); while(SCANF ("%d", n)!=eof N) { for(i=1; i) scanf ("%d",R[i]); b[1] = s[1]; for(i=2; i) {B[i]= b[i-1]; for(j=i-1; j>=1; j--) B[i]= Max (B[i],b[j] +2*min (s[i],s[j])); } for(i=1; i) {Seg[i]. L= B[i]-S[i]; Seg[i]. R= B[i] +S[i]; Seg[i]. LEN=S[i]; } for(i=1; i) { for(j=1; j) { if(Seg[i]. LEN > Seg[j]. LEN Seg[i]. L Seg[j]. R) Seg[j]. R=Seg[i]

]; Ans.push_back (Mintag); Sk.erase (Mintag); Setint>:: Iterator it; while(1) { DoubleMini =Mod; inttag; for(It=sk.begin (); It!=sk.end (); it++) {Point A= p[*it]-Pnow; DoubleAng =Angle (now,a); if(Ang Mini) Mini= ang, Tag = *it; } if(Mini >= Mod) Break; now= p[tag]-Pnow; Pnow=P[tag]; Ans.push_back (tag); Sk.erase (tag); if(Sk.empty ()) Break; } printf ("%d", Ans.size ()); for(i=0; I) printf ("%d", Ans[i]); Puts (""); } retur

. Please bring your computer, and go back to Han dynasty to help them so, the history of the change.Inputthere is no more than the test case.For each test case:The first line is a integer n, meaning that there be N pillars left in E-pang Palace (4 Then N lines follow. Each line contains integers x and y (0 The input ends by N = 0.Outputfor each test case, print the maximum total area of land Zhang Liang and Xiao He could get. If it is impossible for them to build, qualified fences, print "imp".S

RunTime limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)Total submission (s): 101 Accepted Submission (s): 43Problem Descriptionafa is a girl who like runing. Today,he download an app about runing. The app can record the trace of her runing. AFA'll start runing in the park. There is many chairs in the Park,and AFA would start his runing in a chair and end in this chair. Between Chairs,she running in a line.she want the the trace can be a regular triangle or a square or a

Test instructions: There are n individuals to the party, but the condition is that the party location to his home is not more than 2.5, now you need to find the best party location for the most people to party.Analysis:The topic looks very difficult, how to ask scope and then contain points? In fact, a center + RADIUS does not represent a circle.Enumerates the center of a circle with a radius of 2.5 for two points, saves it, and then uses these centers to find the distance of all points to the c

(y) {}6 BOOL operatorConstnode RHS)Const {7 return(x = = rhs.x)? Y rhs.x;8 }9 Ten BOOL operator== (Constnode RHS)Const { One returnx = = Rhs.x y = =Rhs.y; A } - }; - the classLine { - Public: - Node l1,l2; - Line () {} + Line (Node l1,node L2): L1 (L1), L2 (L2) {} - BOOL operatorConstline RHS)Const { + returnL1 = = RHS. L1? L2 RHS. L1; A } at }; - - intMain () { - intN; - //FR ("1.txt"); -CIN >>N; in while(

Topic Link: UVA 319-pendulumNote that height can not be higher than horizontal line, a period is around a point has been circling, a period is to return to the starting point.#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. UVA 319-pendulum (geometry)

Topic Link: UVA 11978-fukushima nuclear BlastTwo points, circle and polygon area intersection. Splits the polygon into an n-part triangle (each two points and the center of the circle) to calculate the direction area. The area of each triangle and circle is discussed in four categories.#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. UVA 11978-fukushima Nuclear Blast (two points + geometry