poj 6206 Apple

Posted 2020-10-08 不积跬步无以至千里，不积小流无以成江海

Apple

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 806    Accepted Submission(s): 267

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

Accepted
Rejected
Rejected

　

只需要输入三个点的坐标即可
p1(x1,y1)  p2(x2,y2)   p3(x3,y3)
A=x1*x1+y1*y1 
B=x2*x2+y2*y2
C=x3*x3+y3*y3
G=(y3-y2)*x1+(y1-y3)*x2+(y2-y1)*x3
X=((B-C)*y1+(C-A)*y2+(A-B)*y3)/(2*G)
Y=((C-B)*x1+(A-C)*x2+(B-A)*x3)/(2*G)

　　

import java.math.BigDecimal;
import java.util.Scanner;


public class Main {

    public static void main(String[] args) {
        Scanner cin=new Scanner(System.in);
        BigDecimal A,B,C,G;
        BigDecimal x1,x2,x3,y1,y2,y3,x0,y0;
        BigDecimal X,Y;
        int t;
        t=cin.nextInt();
        while(t-->0)
        {
            x1=cin.nextBigDecimal();
            y1=cin.nextBigDecimal();
            x2=cin.nextBigDecimal();
            y2=cin.nextBigDecimal();
            x3=cin.nextBigDecimal();
            y3=cin.nextBigDecimal();
            x0=cin.nextBigDecimal();
            y0=cin.nextBigDecimal();
            A=x1.multiply(x1).add(y1.multiply(y1));
            B=x2.multiply(x2).add(y2.multiply(y2));
            C=x3.multiply(x3).add(y3.multiply(y3));
            G=BigDecimal.valueOf(1);
            G=y3.subtract(y2).multiply(x1).add(y1.subtract(y3).multiply(x2)).add(y2.subtract(y1).multiply(x3));
            X=B.subtract(C).multiply(y1).add(C.subtract(A).multiply(y2)).add(A.subtract(B).multiply(y3)).divide(BigDecimal.valueOf(2));
            Y=C.subtract(B).multiply(x1).add(A.subtract(C).multiply(x2)).add(B.subtract(A).multiply(x3)).divide(BigDecimal.valueOf(2));
            x0=x0.multiply(G);
            y0=y0.multiply(G);
            x1=x1.multiply(G);
            y1=y1.multiply(G);
            BigDecimal a=x0.subtract(X).multiply(x0.subtract(X)).add( y0.subtract(Y).multiply(y0.subtract(Y)));
            BigDecimal b=x1.subtract(X).multiply(x1.subtract(X)).add( y1.subtract(Y).multiply(y1.subtract(Y)));
            if(a.compareTo(b)>0)
                System.out.println("Accepted");
            else
                System.out.println("Rejected");
        }
    }
}