zoukankan      html  css  js  c++  java
  • poj 6206 Apple

    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(T30) 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");
            }
        }
    }
    

      

  • 相关阅读:
    什么是线程组,为什么在 Java 中不推荐使用?
    什么是 FutureTask?使用 ExecutorService 启动任务?
    Java 中用到的线程调度算法是什么?
    什么是阻塞队列?阻塞队列的实现原理是什么?如何使用 阻塞队列来实现生产者-消费者模型?
    说说对 SQL 语句优化有哪些方法?(选择几条)
    什么是 Executors 框架?
    Java Concurrency API 中的 Lock 接口(Lock interface) 是什么?对比同步它有什么优势?
    什么是原子操作?在 Java Concurrency API 中有哪些原 子类(atomic classes)?
    Java 中你怎样唤醒一个阻塞的线程?
    你将如何使用 thread dump?你将如何分析 Thread dump?
  • 原文地址:https://www.cnblogs.com/--lr/p/7587275.html
Copyright © 2011-2022 走看看