hdu6206—Apple（计算几何+高精度）

题目链接：传送门

Apple

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 686    Accepted Submission(s): 228

Problem Description

Apple is Taotao's favourite fruit. In his backyard, there are three apple trees with coordinates (x1,y1), (x2,y2), and (x3,y3). Now Taotao is planning to plant a new one, but he is not willing to take these trees too close. He believes that the new apple tree should be outside the circle which the three apple trees that already exist is on. Taotao picked a potential position (x,y) of the new tree. Could you tell him if it is outside the circle or not?

 

Input

The first line contains an integer T, indicating that there are T(T≤30) cases.
In the first line of each case, there are eight integers x1,y1,x2,y2,x3,y3,x,y, as described above.
The absolute values of integers in input are less than or equal to 1,000,000,000,000.
It is guaranteed that, any three of the four positions do not lie on a straight line.

 

Output

For each case, output "Accepted" if the position is outside the circle, or "Rejected" if the position is on or inside the circle.

 

Sample Input

3-2 0 0 -2 2 0 2 -2-2 0 0 -2 2 0 0 2-2 0 0 -2 2 0 1 1

 

Sample Output

AcceptedRejectedRejected

解题思路：高精度求三角形外心

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))； 
因此，x，y为最终结果。

摘自：链接

import java.util.*;import java.util.zip.CRC32;import java.io.*;import java.math.*;public  class Main{public static void main(String[]args){Scanner cin = new Scanner(System.in);int T = cin.nextInt();for( int i = 0 ; i < T ; ++i ){BigDecimal x1,y1,x2,y2,x3,y3,x4,y4;x1 = cin.nextBigDecimal();y1 = cin.nextBigDecimal();x2 = cin.nextBigDecimal();y2 = cin.nextBigDecimal();x3 = cin.nextBigDecimal();y3 = cin.nextBigDecimal();x4 = cin.nextBigDecimal();y4 = cin.nextBigDecimal();BigDecimal A1 = x2.subtract(x1).multiply(BigDecimal.valueOf(2));BigDecimal B1 = y2.subtract(y1).multiply(BigDecimal.valueOf(2));BigDecimal C1 = x2.multiply(x2).add(y2.multiply(y2)).subtract(x1.multiply(x1)).subtract(y1.multiply(y1));BigDecimal A2 = x3.subtract(x2).multiply(BigDecimal.valueOf(2));BigDecimal B2 = y3.subtract(y2).multiply(BigDecimal.valueOf(2));BigDecimal C2 = x3.multiply(x3).add(y3.multiply(y3)).subtract(x2.multiply(x2)).subtract(y2.multiply(y2));BigDecimal x = C1.multiply(B2).subtract(C2.multiply(B1)).divide(A1.multiply(B2).subtract(A2.multiply(B1)));BigDecimal y = A1.multiply(C2).subtract(A2.multiply(C1)).divide(A1.multiply(B2).subtract(A2.multiply(B1)));BigDecimal d1 = x.subtract(x1).multiply(x.subtract(x1)).add(y.subtract(y1).multiply(y.subtract(y1)));BigDecimal d2 = x.subtract(x4).multiply(x.subtract(x4)).add(y.subtract(y4).multiply(y.subtract(y4)));if(d2.compareTo(d1) <= 0) System.out.println("Rejected");else System.out.println("Accepted");}}}