正方形

Time Limit: 1000 ms Memory Limit: 65536 KiB

Problem Description

给出四个点，判断这四个点能否构成一个正方形。

Input

输入的第一行包含一个整数T（T≤30）表示数据组数，每组数据只有一行，包括8个整数x1, y1, x2, y2，x3，y3，x4，y4（数据均在-1000,1000 之间）以逆时针顺序给出四个点的坐标。

Output

每组数据输出一行，如果是正方形，则输出： YES，否则，输出：NO。

Sample Input

0 0 1 0 1 1 0 1

-1 0 0 -2 1 0 2 0

Sample Output

YES

NO

Hint

构建一个与点有关的类，由一些定理我们可以知道，有一个角是直角的菱形就是正方形，所以就可以将问题完全转化为求两点之间距离的问题

package hello;
import java.util.*;
class Point{
	int x, y;
	
	public Point (int x, int y) {
		this.x = x;
		this.y = y;
	}
	
	public double dist(Point p) {
		return (p.x-x)*(p.x-x) + (p.y-y)*(p.y-y);
	}
}
public class Main {  
   
    public static void main(String[] args) {  
        Scanner cin = new Scanner(System.in);  
        int t;
        int x1, y1, x2, y2, x3, y3, x4, y4;
        double d1, d2, d3, d4;
        t = cin.nextInt();
        while(t-- != 0) {
        	x1 = cin.nextInt();
        	y1 = cin.nextInt();
        	x2 = cin.nextInt();
        	y2 = cin.nextInt();
        	x3 = cin.nextInt();
        	y3 = cin.nextInt();
        	x4 = cin.nextInt();
        	y4 = cin.nextInt();
        	Point p = new Point(x1, y1);
        	d1 = p.dist(new Point(x2, y2));
        	p = new Point(x2, y2);
        	d2 = p.dist(new Point(x3, y3));
        	p = new Point(x3, y3);
        	d3 = p.dist(new Point(x4, y4));
        	p = new Point(x1, y1);
        	d4 = p.dist(new Point(x4, y4));
        	double d = p.dist(new Point(x3, y3));
        	//System.out.println(d1+" "+d2+" "+d3+" "+d4+" "+d);
        	if(d1==d2 && d2==d3 && d3==d4 &&d4 == d1 &&d1+d2==d) {
        		System.out.println("YES");
        	}
        	else {
        		System.out.println("NO");
        	}
        }
        cin.close();
    }  
}

正方形