geometry for idiots

materials is different. Then, each raw material is taken out of a certain amount, after melting, mixing, to obtain a new alloy. New alloys of iron and aluminumThe proportion of the tin to the user needs. Now, the user gives the n kinds of alloys they need, and the proportion of Fe-al-sn in each alloy. The company hopes toTo order a minimum variety of raw materials, and use these raw materials to produce all the types of alloys required by the user.InputThe first line is two integers m and n (M,

style= "Display:inline-block; width:0px; Height:2.616em; " > Code ：/**2016-08-04 Morning Author:itakmotto: Today I want to go beyond yesterday's me, tomorrow I want to surpass today's me, to create better code as the goal, constantly surpass oneself. **/#include #include #include #include #include #include #include #include #include using namespace STD;typedef Long LongLL;typedef unsigned Long LongULL;Const intINF =1e9+5;Const intMAXN =1e6+5;Const intMOD =1e9+7;Con

Hdu 4932 Miaomiao #39; s Geometry (brute force enumeration)Miaomiao's GeometryTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Problem DescriptionThere are N point on X-axis. Miaomiao wowould like to cover them ALL by using segments with same length.There are 2 limits:1. A point is convered if there is a segments T, the point is the left end or the right end of T.2. The length of the intersection of any two segments equ

randomly generate a set chart, with the path attribute, the figure size 1*1, find the closest node from the center [0.5, 0.5], and follow the path staining. networkx Examples? Drawing? Random Geometric Graph Copyright NOTICE: Welcome reprint, Reprint please indicate the source http://blog.csdn.net/ztf312/ Python-networkx: Drawing random

adjacent two points //Vgap vertical distance between adjacent two points //hgap*i* (pend[0] //vgap*i* (pend[1] //(pend[0] //(pend[1] //(noslope?0:1) when there is no slope for a line, the horizontal offset is 0Drawpoint ({pw:pw, ph:ph, Color:color, point: [(Hgap*i* (pend[0]])] }); }}Line and polygon can be drawn on the basis of the line:Line://Fold Line//one-dimensional array of PS pointsfunctionDrawPolyline (PS) {if(PS) {//Draw Line for(

= -SinA; SinD= -CosC; Cosd=SinC; Break; default: } //Cross point to draw line linesDrawLine (Point, [Point[0]+c*cosd, Point[1]+c*sind], {color: ' Red '}); DrawLine (Point, [point[0]+c* (-COSD), point[1]+c* (-sind)], {color: ' Red '}); varPoint_v_1 = point[1]-line[1][1]; varpoint_h_1 = point[0]-line[1][0]; varpoint_dist_1 = Math.sqrt (Math.pow (point_v_1,2) +math.pow (point_h_1,2));//point to line up a little distance varPoint_v_0 = point[1]-line[0][1]; varPoint_h_0 = point[0]-line[0][0

, R, Angle, Angleofslope, pop, title) {varNewDot = [Dot[0], dot[1]]; varA = (ANGLEOFSLOPE+ANGLE/2) *math.pi/180; if(POP) {//calculates the new coordinates of the center PointNewdot[0] = dot[0]+10*Math.Cos (a); newdot[1] = dot[1]+10*Math.sin (a); } if(!angleofslope) {Angleofslope= 0; } varAOS = angleofslope*math.pi/180;varAos2 = (angleofslope+angle) *math.pi/180;varPstart = [Newdot[0]+r*math.cos (AOS), Newdot[1]+r*math.sin (AOS)];//the beginning of the arc varPend = [Newdot[0]+r*math

points //DrawArc (Point, quadrant ==1 | | quadrant==4?180:0)-(Quadrant ==2 | | Quadrant==3? ( -1): 1) *math.asin (SinD) *180/math.pi, 0);Drawpoint ({//tangencyPw:3,ph:3,color: ' darkred ', Point:tangentpointa}); Drawpoint ({//tangencyPw:3,ph:3,color: ' darkred ', POINT:TANGENTPOINTB}); //Drawing Auxiliary Arcs //(Quadrant ==1 | | quadrant==4?360:0)DrawArc (point, B, Max, (Quadrant ==1 | | quadrant==4?180:0)-(Quadrant ==2 | | Quadrant==3? ( -1): 1) *math.asin (SinC) *180/math.pi-5);}JS pai

Use the Triangle area to calculate points to the distance of the line and so on ... In fact, the situation of tangency is feasible ... The rest is just like a SDOI2015.#include Compute geometry, dichotomy answer, max flow bzoj1822 [Jsoi2010]frozen Nova freeze Wave

object can consist of one or more PathSegment objects. Each Pathgeomery object uses one and more PathFigure objects that exist in the Paathgeometryfigures collection. Each PathFigure object has one or more PathSegment objects.Important attributes of PathFigure:StartPoint: Specify the starting point of the segmentSegments: A collection of PathSegment objects that are used to draw graphics.IsClosed: If set to true, a straight line connection start and end point is added.Isfilled: If set to true,

than 180, enumerate each point, and find its next target point (with the angle greater than 180 degrees) calculated.1#include 2 #definell Long Long3 #defineN 15054 using namespacestd;5 intn,tp;ll s1,s2;6 structp{7 intx, y;Doubleang;8Poperator- (ConstP b)Const{returnP {x-b.x,y-b.y};}9 BOOL operatorConstP b)Const{returnangB.ang;}Ten }a[n],q[n]; Onell CRS (P a,p b) {return(LL) a.x*b.y-(LL) a.y*b.x;} All solve (intx) { -tp=0; -ll all=1ll* (n1) * (n2) * (n3)/6; the for(intI=1; i){ -

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 (