Uvalive 4986 (search by three points)

There are n vertices in the space. Find a minimum volume cone and pack all vertices. The height and bottom radius of the output cone. The center of the bottom of the cone is (0, 0), and The zcoordinate of all points is greater than or equal to 0.

Problem: Because the cone volume is V = 1/3 * π * r ^ 2 * h, this is a quadratic function, that is, a convex function. You can use a three-point search method to enumerate two heights, find the corresponding minimum r, and compare the two high values to continue the three-point search.

```
# Include
# Include
# Include
Using namespace std; const double PI = acos (-1); const double eps = 1e-9; struct Point {double x, y; Point (double a = 0, double B = 0): x (a), y (B) {}}; typedef Point Vector; double dcmp (double x) {if (fabs (x) <eps) return 0; return x <0? -1: 1;} Vector operator + (const Point & A, const Point & B) {return Vector (. x + B. x,. y + B. y);} Vector operator-(const Point & A, const Point & B) {return Vector (. x-B. x,. y-B. y);} Vector operator * (const Point & A, double a) {return Vector (. x * a,. y * a);} Vector operator/(const Point & A, double a) {return Vector (. x/a,. y/a);} double Cross (const Vector & A, const Vector & B) {return. x * B. y-. y * B. x;} double Dot (const Vector & A, const Vector & B) {return. x * B. x +. y * B. y;} double Length (const Vector & A) {return sqrt (Dot (A, A);} bool operator <(const Point & A, const Point & B) {return. x <B. x | (. x = B. x &. y <B. y);} bool operator = (const Point & A, const Point & B) {return. x = B. x &. y = B. y;} const int N = 10005; Point P [N], HP [N]; int n; double solve (double h) {double r =-1; // (P [I]. x-0, P [I]. y-h) (r-0, 0-h) Coaxial for (int I = 0; I <n; I ++) if (P [I]. x * h/(h-P [I]. y)> r) r = P [I]. x * h/(h-P [I]. y); return r;} int main () {while (scanf (% d, & n) = 1) {double x, y, z, l = 0, r = 1e9; for (int I = 0; I <n; I ++) {scanf (% lf, & x, & y, & z ); double d = sqrt (x * x + y * y); P [I]. x = d; P [I]. y = z; l = max (l, z);} while (dcmp (r-l)> 0) {double h1 = (l + r)/2; double h2 = (h1 + r)/2; double r1 = solve (h1); double r2 = solve (h2 ); if (r1 * r1 * h1 <r2 * r2 * h2) r = h2; else l = h1;} printf (%. 3lf %. 3lf, l, solve (l);} return 0 ;}
```