uvalive 4986(三分查找)

uvalive 4986(三分查找)

題意：空間內有n個點，求一個最小體積的圓錐把所有點包進去。輸出圓錐的高和底面半徑。圓錐的底面圓心在(0,0)，所有點的z坐標都大於等於0。
題解：因為圓錐體積是 V = 1/3 * π * r^2 * h ，這是一個二次函數，也就是個凸性函數，可以用三分查找的方式枚舉兩個高，然後找到對應的最小的r，比對兩個高得到的體積繼續三分查找。

#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) 共線 
    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;
}