在二维平面中，给出了四个点。该算法将检查四个点是否形成正方形。
检查正方形我们必须匹配这些条件-
给定点形成的所有四个边都相同。
所有两个连接侧都是直角的。
Input:
Four points {(20, 10), (10, 20), (20, 20), (10, 10)}
Output:
点正在形成一个正方形。isFormingSquare(p1, p2, p3, p4)
在此过程中，我们将使用方法squareDist(p1,p2)，它将返回两个给定点的平方距离。
输入： 四点。
输出：给定点形成正方形时为真。
Begin dist12 := squareDist(p1, p2) dist13 := squareDist(p1, p3) dist14 := squareDist(p1, p4) if dist12 = dist13 and 2*dist12 = dist14, then dist := squareDist(p2, p4) return true when dist = squareDist(p3, p4) and dist = dist12 if dist13 = dist14 and 2*dist13 = dist12, then dist := squareDist(p2, p3) return true when dist = squareDist(p2, p4) and dist = dist13 if dist12 = dist14 and 2*dist12 = dist13, then dist := squareDist(p2, p3) return true when dist = squareDist(p3, p4) and dist = dist12 return false End
#include<iostream>
using namespace std;
struct Point {
int x, y;
};
int squareDist(Point p, Point q) {
return (p.x - q.x)*(p.x - q.x) + (p.y - q.y)*(p.y - q.y);
}
bool isSquare(Point p1, Point p2, Point p3, Point p4) { //check four points are forming square or not
int dist12 = squareDist(p1, p2); // distance from p1 to p2
int dist13 = squareDist(p1, p3); // distance from p1 to p3
int dist14 = squareDist(p1, p4); // distance from p1 to p4
//当p1-p2和p1-p3的长度相同时，(p1-p4)的平方= 2 *(p1-p2)
if (dist12 == dist13 && 2*dist12 == dist14) {
int dist = squareDist(p2, p4);
return (dist == squareDist(p3, p4) && dist == dist12);
}
//所有其他组合的条件相同
if (dist13 == dist14 && 2*dist13 == dist12) {
int dist = squareDist(p2, p3);
return (dist == squareDist(p2, p4) && dist == dist13);
}
if (dist12 == dist14 && 2*dist12 == dist13) {
int dist = squareDist(p2, p3);
return (dist == squareDist(p3, p4) && dist == dist12);
}
return false;
}
int main() {
Point p1 = {20, 10}, p2 = {10, 20}, p3 = {20, 20}, p4 = {10, 10};
if(isSquare(p1, p2, p3, p4))
cout << "点正在形成一个正方形。";
else
cout << "Points are not forming a square";
}输出结果
点正在形成一个正方形。