hdu 5017 Ellipsoid(三分)

题目链接：hdu 5017 Ellipsoid

题目大意：给定一个面的方程，问在面上距离原点的最小值。

解题思路：三分套三分，先三分x，对于每个x，三分y，求出的最优解作为当前x的值。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
const double INF = 10000;
const double eps = 1e-9;
double a, b, c, d, e, f;

double get(double A, double B, double C) {
    if (B * B - 4 * A * C < eps)
        return INF;
    return (sqrt(B * B - 4 * A * C) - B) / (2 * A);
}

double func (double x, double y) {
    double z = get(c, e * x + d * y, a * x * x + b * y * y + f * x * y - 1);
    return x * x + y * y + z * z;
}

double search (double x) {
    double l = -INF, r = INF;
    for (int i = 0; i < 200; i++) {
        double ll = l + (r - l) / 3;
        double rr = r - (r - l) / 3;
        if (func(x, ll) > func(x, rr))
            l = ll;
        else
            r = rr;
    }
    return func(x, l);
}

double solve (double l, double r) {
    for (int i = 0; i < 200; i++) {
        double ll = l + (r - l) / 3;
        double rr = r - (r - l) / 3;
        if (search(ll) > search(rr))
            l = ll;
        else
            r = rr;
    }
    return sqrt(search(l));
}

int main () {
    while (scanf("%lf%lf%lf%lf%lf%lf", &a, &b, &c, &d, &e, &f) == 6) {
        printf("%.5lf\n", solve(-INF, INF));
    }
    return 0;
}