HDU 6205[计算几何,JAVA]

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

题意:

  给出不共线的三个点,和一个点(x,y),然后判断(x,y)在不在这三个点组成的圆外。

题解:

  咋一看很简单,实际也很简单,但是坐标都很大,会爆long double,怎么办?只有用java了。

公式:

a = ((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)));
b = ((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 - a) * (x1 - a) + (y1 - b) * (y1 - b);

  

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

public class Main {

    public static void main(String[] args) {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        BigDecimal x1, y1, x2, y2, x3, y3, x, y;
        int T = cin.nextInt();
        for (int k = 1; k <= T; k++) {
            x1 = cin.nextBigDecimal();
            y1 = cin.nextBigDecimal();
            x2 = cin.nextBigDecimal();
            y2 = cin.nextBigDecimal();
            x3 = cin.nextBigDecimal();
            y3 = cin.nextBigDecimal();
            x = cin.nextBigDecimal();
            y = cin.nextBigDecimal();

            BigDecimal t1 = y2.subtract(y1);
            BigDecimal t2 = y3.multiply(y3);
            t2 = t2.subtract(y1.multiply(y1));
            t2 = t2.add(x3.multiply(x3));
            t2 = t2.subtract(x1.multiply(x1));
            BigDecimal t3 = y3.subtract(y1);
            BigDecimal t4 = y2.multiply(y2);
            t4 = t4.subtract(y1.multiply(y1));
            t4 = t4.add(x2.multiply(x2));
            t4 = t4.subtract(x1.multiply(x1));
            BigDecimal t5 = (x3.subtract(x1)).multiply(y2.subtract(y1));
            t5 = t5.subtract(x2.subtract(x1).multiply(y3.subtract(y1)));
            t5 = t5.multiply(BigDecimal.valueOf(2.0));
            BigDecimal a = ((t1.multiply(t2)).subtract(t3.multiply(t4))).divide(t5);


            t1 = x2.subtract(x1);
            t3 = x3.subtract(x1);
            t5 = (y3.subtract(y1)).multiply(x2.subtract(x1));
            t5 = t5.subtract(y2.subtract(y1).multiply(x3.subtract(x1)));
            t5 = t5.multiply(BigDecimal.valueOf(2.0));
            BigDecimal b = ((t1.multiply(t2)).subtract(t3.multiply(t4))).divide(t5);
            BigDecimal r = (x1.subtract(a)).multiply(x1.subtract(a));
            r = r.add((y1.subtract(b)).multiply(y1.subtract(b)));

            BigDecimal R = (x.subtract(a)).multiply(x.subtract(a));
            R = R.add((y.subtract(b)).multiply(y.subtract(b)));
            if(R.compareTo(r) > 0) {
                System.out.println("Accepted");
            }
            else {
                System.out.println("Rejected");
            }
        }
    }
}

  

posted @ 2017-09-18 10:25  _Mickey  阅读(162)  评论(0编辑  收藏  举报