uvalive 4986(三分尋找)

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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座標都大於等於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，可以用三分尋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尋找。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>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\n", l, solve(l));       }    return 0;}

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uvalive 4986(三分尋找)