题意：空间内有n个点，求一个最小体积的圆锥把所有点包进去。输出圆锥的高和底面半径。圆锥的底面圆心在(0,0)，所有点的z坐标都大于等于0。
题解：因为圆锥体积是 V = 1/3 * π * r^2 * h ，这是一个二次函数，也就是个凸性函数，可以用三分查找的方式枚举两个高，然后找到对应的最小的r，比对两个高得到的体积继续三分查找。

#include 
   
     #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(A.x + B.x, A.y + B.y); } Vector operator - (const Point& A, const Point& B) { return Vector(A.x - B.x, A.y - B.y); } Vector operator * (const Point& A, double a) { return Vector(A.x * a, A.y * a); } Vector operator / (const Point& A, double a) { return Vector(A.x / a, A.y / a); } double Cross(const Vector& A, const Vector& B) { return A.x * B.y - A.y * B.x; } double Dot(const Vector& A, const Vector& B) { return A.x * B.x + A.y * B.y; } double Length(const Vector& A) { return sqrt(Dot(A, A)); } bool operator < (const Point& A, const Point& B) { return A.x < B.x || (A.x == B.x && A.y < B.y); } bool operator == (const Point& A, const Point& B) { return A.x == B.x && A.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) 共线 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%lf%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; }
      
     
    
   

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