HDU 1700 points on cycle (geometric vector rotation)

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1700

 

Question:

A two-dimensional plane with the center of a circle on the origin. Given the point A on the circle, find the other two points B, C, B, and C on the circle, and the circumference of the Triangle ABC is the longest.

 

Solution:

I remember giving a theorem in elementary school. the circumference of the positive polygon in the garden is the longest. I will not prove this theorem.

This is a triangle. When the triangle is a positive triangle, the circumference is the longest.

Because the center of the center is at the origin, I rotate the vector (x, y) around the origin 120 degrees clockwise and 120 degrees clockwise. The given vertex A can be considered as the (x, y) vector.

Vector rotation has formulas (p256, a classic training guide for algorithm competitions ).

 

AC code:

 1 #include<cmath> 2 #include<cstdio> 3 #include<algorithm> 4 using namespace std; 5  6 const double PI = acos(-1.0); 7  8 struct Point{ 9     double x, y;10 11     Point(double x = 0, double y = 0): x(x), y(y){}12 13     void scan(){14         scanf("%lf%lf", &x, &y);15     }16 17     void print(){18         printf("%.3lf %.3lf", x, y);19     }20 21     bool operator < (const Point &other){22         return y < other.y || (y == other.y && x < other.x);23     }24 };25 26 typedef Point Vector;27 28 Vector rotate(Vector A, double rad){//向量旋转公式29     return Vector(A.x * cos(rad) - A.y * sin(rad), A.y * cos(rad) + A.x * sin(rad));30 }31 32 int main(){33     int t;34     Point p[3];35     scanf("%d", &t);36     while(t--){37         p[0].scan();38 39         p[1] = rotate(p[0], PI * 2 / 3);//逆时针旋转120度40         p[2] = rotate(p[0], -PI * 2 / 3);//顺时针旋转120度41 42         if(p[2] < p[1]) swap(p[1], p[2]);//按题目要求输出43 44         p[1].print(); putchar(‘ ‘);45         p[2].print(); putchar(‘\n‘);46     }47     return 0;48 }

 

HDU 1700 points on cycle (geometric vector rotation)