HDU 6206 Apple (high-precision && computational Geometry && Three-point construction Circle for center radius)

Test instructions: give four points, ask you whether the fourth point in the first three points constitute the circle, if the output of "Accepted" outside the circle, otherwise output "rejected", the title guarantee the first three points are not in a straight line.

Analysis: A simple computational geometry problem, if you can know the center and radius (radius) and the distance between the fourth and center points (Distance), we can determine whether the fourth point is outside the circle, such as Distance > Radius is outside the circle. The center and radius of the three-point structure circle are the formulas that can be deduced.

(Refer to ==> http://blog.csdn.net/death10001/article/details/77389128)

The setting has

According to the distance between two points formula R2 = (x-x0) 2 + (Y-Y0) 2 We will just the X0 algebra and X (any one of three points) to go in-pass a little to get

[2*x* (Bc-ad) + (DE-BF)/2 (BC-AD)]2 + [2*y* (bc-ad) + (AF-CE)/2 (BC-AD)] 2

Find the denominator is the same, can-pass. Then, or according to two-point spacing we can use the same method to get the fourth point and the distance between the center and the denominator is also 2 (BC-AD), so the process of operation is not to divide operations, as long as the comparison of two molecules can be, high-precision problems can only be solved with BigInteger.

ImportJava.util.Scanner;ImportJava.util.*;Importjava.math.*;ImportJava.io.*;classpo{BigInteger x, y;} Public classmain{ Public Static BooleanSolve (po p1, po p2, PO P3, po p) {//-------------------------------------------------------------------------------------BigInteger a=p1.x.subtract (p2.x); BigInteger b=p1.y.subtract (P2.Y); BigInteger C=p1.x.subtract (p3.x); BigInteger D=p1.y.subtract (P3.Y); BigInteger e=((p1.x). Multiply (p1.x)). Subtract ((p2.x). Multiply (p2.x)). Subtract (((P2.Y). Multiply (P2.Y)). Subtract (((p1.y        ). Multiply (P1.Y))); BigInteger F=((p1.x). Multiply (p1.x)). Subtract ((p3.x). Multiply (p3.x)). Subtract (((P3.Y). Multiply (P3.Y)). Subtract (((p1.y        ). Multiply (P1.Y))); BigInteger=NewBigInteger ("2"); //-------------------------------------------------------------------------------------BigInteger TMP1=((p1.x). Multiply ((b.multiply (c)). Subtract ((a.multiply (d)))). Add ((Multiply (e)). D.multiply (        (B.multiply (f)))); TMP1=tmp1.multiply (TMP1); BigInteger TMP2=((P1.Y). Multiply ((b.multiply (c)). Subtract ((a.multiply (d)))). Add ((Multiply (f)). A.multiply (        (C.multiply (e)))); TMP2=tmp2.multiply (TMP2); BigInteger R= Tmp1.add (TMP2);///Calculate the "length" of the radius (actually the molecule of algebraic type)//-------------------------------------------------------------------------------------TMP1=((p.x). Multiply ((b.multiply (c)). Subtract ((a.multiply (d)))). Add ((Multiply (e)). D.multiply ((        B.multiply (f))); TMP1=tmp1.multiply (TMP1); TMP2=((P.Y). Multiply ((b.multiply (c)). Subtract ((a.multiply (d)))). Add ((Multiply (f)). A.multiply ((        C.multiply (e))); TMP2=tmp2.multiply (TMP2); BigInteger Dis= Tmp1.add (TMP2);///Calculates the "distance" from the center to the fourth point        if(R.compareto (Dis) < 0)return true; Else return false; }     Public Static voidMain (string[] args)throwsexception{Scanner Cin=NewScanner (NewBufferedinputstream (system.in)); intNcase =Cin.nextint ();  for(intT=1; t<=ncase; t++) {PO P1=NewPO (); PO P2 =NewPO (); PO P3=NewPO (); PO aim=NewPO (); p1.x= Cin.nextbiginteger (); P1.Y =Cin.nextbiginteger (); p2.x= Cin.nextbiginteger (); P2.Y =Cin.nextbiginteger (); p3.x= Cin.nextbiginteger (); P3.Y =Cin.nextbiginteger (); Aim.x= Cin.nextbiginteger (); AIM.Y =Cin.nextbiginteger (); Booleanres =Solve (P1, P2, p3, AIM); if(RES) System.out.println ("Accepted"); ElseSystem.out.println ("Rejected"); }    }}

View Code

Blindness: There is actually a simpler way to calculate the square of the center (x0, y0) and radius directly with a more straightforward formula r^2

X0= ((y2-y1) * (y3*y3-y1*y1+x3*x3-x1*x1)-(y3-y1) * (Y2*Y2-Y1*Y1+X2*X2-X1*X1))/(2.0* ((x3-x1) * (y2-y1)-(x2-x1) * (y3-y1) ));

Y0= ((x2-x1) * (x3*x3-x1*x1+y3*y3-y1*y1)-(x3-x1) * (x2*x2-x1*x1+y2*y2-y1*y1))/(2.0* ((y3-y1) * (x2-x1)-(y2-y1) * (X3-X1) ));

R^2= (x1-x0) * (x1-x0) + (y1-y0) * (Y1-Y0);

It involves division, that can be solved with BigDecimal, reference ==> http://blog.csdn.net/cillyb/article/details/78012069

HDU 6206 Apple (high-precision && computational Geometry && Three-point construction Circle for center radius)