computational geometry book

Getcenter [Point points[],intN) { the Point Res; * Doublearea,areamulx,areamuly; $Area =0.0; Areamulx =0.0; Areamuly =0.0;Panax Notoginseng for(intI=1; i1; i++){ -VECTOR V1 = points[i]-points[0]; theVECTOR v2 = points[i+1]-points[i];//if Fabs will lose precision after + DoubleTarea = V1 * V2/2.0; AArea + =Tarea; theAreamulx + = (points[0]. X+points[i]. x+points[i+1]. X) *Tarea; +Areamuly + = (points[0]. Y+points[i]. y+points[i+1]. Y) *Tarea; - } $Res. X = Areamulx/(3.0*Are

Title Address: POJ 1905Double-enumerate H, and then determine if the arc length meets the criteria. Focus on the accuracy of the problem, the specific look at the code bar.#include POJ 1905 Expanding Rods (two points + computational geometry + precision Processing)

Basic and online solution the same idea, put several cases blablabla. It's WA.Then add an EPS to the answer. Obviously this question did not write SPJ. The accuracy of the cardCode:#include POJ 2826 an easy problem?! (Computational geometry)

Cross-product utilization inferred point segments left or right, and then you can 2 minutesCode:#include POJ 2318 TOYS (computational geometry)

//=========================================================================DoubleCP (V a,v b) {returna.px*b.py-a.py*b.px;}//Cross productDoubleDP (V a,v b) {returnA.PX*B.PX + a.py*b.py;}//dot ProductDoubleDIS (V a,v b) {returnsqrt ((a.px-b.px) * (A.PX-B.PX) + (a.py-b.py) * (a.py-b.py));}//disdance//=========================================================================BOOLPOL (V q,l p) {/*double x1 = p.p1.px,x2 = P.p2.px,y1 = P.p1.py,y2 = p.p2.py, x0 = Q.px,y0 = q.py; if ((!P.P0) (!) ( X

line and the slope of a large line, the size of the two X coordinate to compare, then you can ~Code://Qscqesze#include #include#include#include#include#include#includeSet>#include#include#include#include#include#includetypedefLong Longll;using namespacestd;//freopen ("d.in", "R", stdin);//freopen ("D.out", "w", stdout);#defineSspeed ios_base::sync_with_stdio (0); Cin.tie (0)#defineMAXN 50001#defineMoD 10007#defineEPS 1e-9//const int INF=0X7FFFFFFF; //infinitely LargeConst intinf=0x3f3f3f3f;/**/

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

HDU 4173Test instructions: Known n (nIdeas:Consider the possible location of the party, and to invite as many players as possible, just consider where the party is located at the midpoint of a two-bit residence line or the location of a contestant's residence, as this is where the largest number of participants is likely to be located.If the maximum number of participants can be obtained from other locations, then the midpoint or the contestant's residence will be available. Ok~, try to draw.So,

, the robot goes backwards from the 1th-cell order to the N-grid and needs to go out of the N-grid. The robot has an initial energy, each lattice corresponds to an integer a[i], which represents the energy value of the lattice. If A[i] > 0, the robot goes to this lattice can get a[i] energy, if a[i] For example:n = 5. {1,-2,-1,3,4} requires a minimum of 2 initial energy to go from number 1th to Grid 5th. The energy changes on the way are as follows 3 1 0 3 7.Input Line 1:1 number n, indicating t

Topic: Given a polygon, the number of axes of symmetryI x this is how to think KMP ...The Edge word Fu Hua is then found in the center of the polygon and represented by the edge-angle-edge of the triangle with the center.Sweep the side clockwise over and then multiply to the length of 2n-1 and then counterclockwise sweep again, the number of times that appear in the clockwise that is the number of axesWith the KMP algorithm, you can prove yourself yy.The accuracy of human card ...#include Bzoj 1

] +Weight[i]; } intAns =-1; //The last part of the Tocheck is J to i+f.//The core idea is that if J-I is the part that pulls down the whole tocheck, don't.//The core idea above is very much like the two-part answer to the idea that there is more than or equal to the F term and greater than 0. for(inti =0, j =0; I ) { if(i > J (Presum[i]-presum[j]) * (i + f-j) j)) J= i;//if the average of the sub-segments between I and J is less than the average between I and I+f, the sub-segme

Topic link Analysis See through the problem Considering every line segment that makes up the village, it is clear that we will see it in the half plane above the line where the line is located. Therefore, watchtower must be in the intersection of the half plane above the line where all the line segments of the village are formed, in order to From the top of watchtower, you can see anywhere in H village. Therefore, this problem is to seek the shortest distance from the ground of the village to

Topic link Pigs can ' t take a sudden turn Time limit:1000 Ms Memory limit:65536 KB problem Descrpition You maybe has aced a question about computational Geometry,which describes the dogs ' journey and calculate the shortest D Istance between them. Now, I provide a easier problem about the pigs. Heir Path is half-lines. When they start running, they won ' t stop. Why? Because they can ' t take a sudden turn