Java练习 SDUT-2444_正方形

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

2
0 0 1 0 1 1 0 1
-1 0 0 -2 1 0 2 0

Sample Output

YES
NO

题解：根据正方形的判定公式，对角线相等且垂直，来判断是不是正方形。

import java.util.*;

public class Main {
	public static void main(String[] agrs)
	{
		Scanner cin = new Scanner(System.in);
		point a,b,c,d;
		int t;
		t = cin.nextInt();
		while(t-->0)
		{
			a = new point();
			b = new point();
			c = new point();
			d = new point();
			a.x = cin.nextInt();
			a.y = cin.nextInt();
			b.x = cin.nextInt();
			b.y = cin.nextInt();
			c.x = cin.nextInt();
			c.y = cin.nextInt();
			d.x = cin.nextInt();
			d.y = cin.nextInt();
			if(judge(a,b,c,d)==1)
				System.out.println("YES");
			else
				System.out.println("NO");
		}
		cin.close();
	}
	static int judge(point a,point b,point c,point d)
	{
		int x,y;
		x = (a.x-c.x) * (a.x-c.x) + (a.y-c.y) * (a.y-c.y);
		y = (b.x-d.x) * (b.x-d.x) + (b.y-d.y) * (b.y-d.y);
		if(x==y&&((a.x-c.x) * (b.x-d.x)==-((a.y-c.y) * (b.y-d.y))))
			return 1;
		return 0;
	}
}

class point
{
	int x,y;
}