HDU 6206 Apple

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6206

判断给定一点是否在三角形外接圆内。

给定三角形三个顶点的坐标，如何求三角形的外心的坐标呢？

知乎链接：https://www.zhihu.com/question/40422123/answer/86514178

例如 ：给定a（x1,y1） b（x2,y2） c（x3,y3）求外接圆心坐标O（x，y）

1. 首先，外接圆的圆心是三角形三条边的垂直平分线的交点，我们根据圆心到顶点的距离相等，可以列出以下方程：
(x1-x)*(x1-x)+(y1-y)*(y1-y)=(x2-x)*(x2-x)+(y2-y)*(y2-y);
(x2-x)*(x2-x)+(y2-y)*(y2-y)=(x3-x)*(x3-x)+(y3-y)*(y3-y);

2.化简得到：
2*(x2-x1)*x+2*(y2-y1)y=x2^2+y2^2-x1^2-y1^2;
2*(x3-x2)*x+2*(y3-y2)y=x3^2+y3^2-x2^2-y2^2;

令:A1=2*(x2-x1)；
B1=2*(y2-y1)；
C1=x2^2+y2^2-x1^2-y1^2;
A2=2*(x3-x2)；
B2=2*(y3-y2)；
C2=x3^2+y3^2-x2^2-y2^2;
即:A1*x+B1y=C1;
A2*x+B2y=C2;

3.最后根据克拉默法则：
x=((C1*B2)-(C2*B1))/((A1*B2)-(A2*B1))；
y=((A1*C2)-(A2*C1))/((A1*B2)-(A2*B1))；

所以x,y为圆心坐标。

然后最后r^2  = (x1-x)*(x1-x)+(y1-y)*(y1-y)就好了。

C++卡double，换成Java写，用BigDecimal写。

 import java.io.*;
 import java.math.BigDecimal;
 import java.math.BigInteger;
 import java.util.*; 

 public class Main {

     final static BigInteger HUNDRED = BigInteger.valueOf(100);  

     //叉乘求面积
    /* private static BigDecimal TriangleArea(POINT pi,POINT pj,POINT pk)
     {
         BigDecimal num1 =  (pj.x.subtract(pi.x)).multiply(pk.y.subtract(pi.y));
         BigDecimal num2 =  (pk.x.subtract(pi.x)).multiply(pj.y.subtract(pi.y));
         BigDecimal num3 = (num1.subtract(num2)).abs();
         BigDecimal two = new BigDecimal("2.0");
         return num3.divide(two);
     }*/
     //平方
     private static BigDecimal pinfang(BigDecimal x)
     {
         return x.multiply(x);
     }

     //求长度
     private static BigDecimal Dis(POINT a,POINT b)
     {
         return pinfang(a.x.subtract(b.x)).add(pinfang(a.y.subtract(b.y)));
     }

     public static void main(String[] args) {
         // TODO 自动生成的方法存根
         Scanner cin=new Scanner(System.in);
         int T = cin.nextInt();
         for(int i=1;i<=T;i++)
         {
             POINT A = new POINT();
             POINT B = new POINT();
             POINT C = new POINT();
             POINT D = new POINT();

             A.x = cin.nextBigDecimal();
             A.y = cin.nextBigDecimal();
             B.x = cin.nextBigDecimal();
             B.y = cin.nextBigDecimal();
             C.x = cin.nextBigDecimal();
             C.y = cin.nextBigDecimal();
             D.x = cin.nextBigDecimal();
             D.y = cin.nextBigDecimal();

             BigDecimal two = new BigDecimal("2.0");
             BigDecimal four = new BigDecimal("4.0");

             //求圆心
             BigDecimal c1 = ((pinfang(A.x).add(pinfang(A.y))).subtract(pinfang(B.x).add(pinfang(B.y)))).divide(two);
             BigDecimal c2 = ((pinfang(A.x).add(pinfang(A.y))).subtract(pinfang(C.x).add(pinfang(C.y)))).divide(two);

             POINT center = new POINT();
             center.x = (c1.multiply((A.y.subtract(C.y)))).subtract(c2.multiply(A.y.subtract(B.y)));
             BigDecimal tmp = ((A.x.subtract(B.x)).multiply(A.y.subtract(C.y))).subtract((A.x.subtract(C.x)).
                     multiply(A.y.subtract(B.y)));
             center.x = center.x.divide(tmp);
             center.y = (c1.multiply((A.x.subtract(C.x)))).subtract(c2.multiply(A.x.subtract(B.x)));
             tmp = ((A.y.subtract(B.y)).multiply(A.x.subtract(C.x))).subtract((A.y.subtract(C.y)).
                     multiply(A.x.subtract(B.x)));
             center.y = center.y.divide(tmp);
             //求半径
             BigDecimal r2 = pinfang(A.x.subtract(center.x)).add(pinfang(A.y.subtract(center.y)));
             //圆心到D点距离
             BigDecimal tempr = pinfang(D.x.subtract(center.x)).add(pinfang(D.y.subtract(center.y)));

             if(tempr.compareTo(r2)<=0)
             {
                 System.out.println ("Rejected");
             }
             else
             {
                 System.out.println ("Accepted");
             }
         }
     }
 }

 class POINT
 {

     BigDecimal x;
     BigDecimal y;
 }

 /*
  41
  0 0 1 1 2 0 2 0
 -2 0 0 -2 2 0 2 -2
 -2 0 0 -2 2 0 0 2
 -2 0 0 -2 2 0 1 1
  */