PAT 三角形判断 Java版本

/*题目描述：给定平面上任意三个点的坐标(x1,y1)、(x2,y2)、(x3,y3)，检验它们能否构成三角形。输入格式：输入在一行中顺序给出6个[-100, 100]范围内的数字，即3个点的坐标x1, y1, x2, y2, x3, y3。输出格式：若这3个点不能构成三角形，则在一行中输出“Impossible”；若可以，则在一行中输出该三角形的周长和面积，格式为“L = 周长, A = 面积”，输出到小数点后2位。输入样例1：

4 5 6 9 7 8

输出样例1：

L = 10.13, A = 3.00

输入样例2：

4 6 8 12 12 18

输出样例2：

Impossible

*/

/*

下面是我写的Java解法，比较麻烦，但是正确的。

看了网上的方法，发现更简单的做法是利用三角形的面积公式：S*S=p*(p-a)*(p-b)*(p-c),其中p=(a+b+c)/2.

*/

import java.util.*;import java.lang.*;import java.text.*;public class Main {class Point{double x=0.0,y=0.0;}public Point createPoint(){return new Point();}public double CalculateArea(Point point1, Point point2, Point point3){double area = 0.0;double area1 = (point1.y+point2.y)*Math.abs(point2.x-point1.x)/2.0;double area2 = (point3.y+point2.y)*Math.abs(point2.x-point3.x)/2.0;double area3 = (point1.y+point3.y)*Math.abs(point3.x-point1.x)/2.0;area = Math.abs(area1+area2-area3);return area;}public static void main(String[] args){Main obj = new Main();Scanner in = new Scanner(System.in);Main.Point point1 = obj.createPoint();Main.Point point2 = obj.createPoint();Main.Point point3 = obj.createPoint();point1.x = in.nextDouble();point1.y = in.nextDouble();point2.x = in.nextDouble();point2.y = in.nextDouble();point3.x = in.nextDouble();point3.y = in.nextDouble();double d1 = (point1.x-point2.x)*(point1.x-point2.x)+(point1.y-point2.y)*(point1.y-point2.y);double d2 = (point1.x-point3.x)*(point1.x-point3.x)+(point1.y-point3.y)*(point1.y-point3.y);double d3 = (point2.x-point3.x)*(point2.x-point3.x)+(point2.y-point3.y)*(point2.y-point3.y);if(Math.sqrt(d1)+Math.sqrt(d2)>Math.sqrt(d3) && Math.sqrt(d1)+Math.sqrt(d3)>Math.sqrt(d2) &&Math.sqrt(d2)+Math.sqrt(d3)>Math.sqrt(d1)){DecimalFormat df = new DecimalFormat("#0.00");double L = Math.sqrt(d1)+Math.sqrt(d2)+Math.sqrt(d3);System.out.print("L = "+df.format(L)+", ");double[] arrX = {point1.x,point2.x,point3.x};double[] arrY = {point1.y,point2.y,point3.y};Arrays.sort(arrX);Arrays.sort(arrY);Main.Point zero = obj.createPoint();zero.x = arrX[0];zero.y = arrY[0];point1.x = point1.x - zero.x;point1.y = point1.y - zero.y;point2.x = point2.x - zero.x;point2.y = point2.y - zero.y;point3.x = point3.x - zero.x;point3.y = point3.y - zero.y;double area = 0.0;if((point1.x-point2.x)*(point1.x-point3.x)<=0){area = obj.CalculateArea(point2, point1, point3);}else if((point2.x-point1.x)*(point2.x-point3.x)<=0){area = obj.CalculateArea(point1, point2, point3);}else{area = obj.CalculateArea(point1, point3, point2);}System.out.print("A = "+df.format(area)+"\n");}else{System.out.println("Impossible");}}}