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
第一种思路:
判断四个边相等,加一个直角
import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner reader=new Scanner(System.in);
int T=reader.nextInt();
for(int i=1;i<=T;i++)
{
int x1=reader.nextInt();
int y1=reader.nextInt();
int x2=reader.nextInt();
int y2=reader.nextInt();
int x3=reader.nextInt();
int y3=reader.nextInt();
int x4=reader.nextInt();
int y4=reader.nextInt();
Point p1=new Point(x1, y1);
Point p2=new Point(x2, y2);
Point p3=new Point(x3, y3);
Point p4=new Point(x4, y4);
int l1=p1.dis(p2);
int l2=p2.dis(p3);
int l3=p3.dis(p4);
int l4=p4.dis(p1);
int m=p1.dis(p3);
if(l1==l2&&l2==l3&&l3==l4&&l1+l2==m)
{
System.out.println("YES");
}
else
{
System.out.println("NO");
}
}
reader.close();
}
}
class Point
{
int x;
int y;
public Point(int x,int y)
{
this.x=x;
this.y=y;
}
public int dis(Point p)
{
int distance=(x-p.x)*(x-p.x)+(y-p.y)*(y-p.y);
return distance;
}
}
第二种方法:
判断四个边相等加对角线相等
import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner reader=new Scanner(System.in);
int T=reader.nextInt();
for(int i=1;i<=T;i++)
{
int x1=reader.nextInt();
int y1=reader.nextInt();
int x2=reader.nextInt();
int y2=reader.nextInt();
int x3=reader.nextInt();
int y3=reader.nextInt();
int x4=reader.nextInt();
int y4=reader.nextInt();
Point p1=new Point(x1, y1);
Point p2=new Point(x2, y2);
Point p3=new Point(x3, y3);
Point p4=new Point(x4, y4);
int l1=p1.dis(p2);
int l2=p2.dis(p3);
int l3=p3.dis(p4);
int l4=p4.dis(p1);
int m1=p1.dis(p3);
int m2=p2.dis(p4);
if(l1==l2&&l2==l3&&l3==l4&&m1==m2)
{
System.out.println("YES");
}
else
{
System.out.println("NO");
}
}
reader.close();
}
}
class Point
{
int x;
int y;
public Point(int x,int y)
{
this.x=x;
this.y=y;
}
public int dis(Point p)
{
int distance=(x-p.x)*(x-p.x)+(y-p.y)*(y-p.y);
return distance;
}
}