HDU 3756 Dome of Circus(计算几何 + 三分)_acm hdu dome of circus-优快云博客

本文链接：https://blog.youkuaiyun.com/hnust_Derker/article/details/52516254

本文介绍了一种解决三维空间中给定点集最小圆锥覆盖问题的方法。通过将三维问题转化为二维平面上的点被三角形包围的问题，并采用三分法求解最优斜率，最终确定最小覆盖圆锥。

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题意：给定三维空间中n个点坐标，必在xOy平面之上，求一个最小的圆锥能包含所有顶点，点可以刚好在面上

思路：转换为平面上的点被一个三角形包围，Y坐标为原来的z坐标，X坐标为原来到z轴的距离，然后三分求解斜率，注意精度问题

#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<string>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#include<iostream>
typedef long long ll;
const int maxn = 1e4 + 10;
const double eps = 1e-7;
const ll INF = 1e18;
using namespace std;

double x, y, z;
double X[maxn], Y[maxn];
int n, k, T;

int main()
{
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for(int i = 0; i < n; i++) {
            scanf("%lf %lf %lf", &x, &y, &z);
            X[i] = sqrt(x * x + y * y);
            Y[i] = z;
        }
        double res = INF, l = -INF, r = 0;
        double R, H;
        while(l < r) {
            double res1 = 0, res2 = 0;
            double rr1, hh1, rr2, hh2;
            double mid = (l + r) / 2, mmid = (mid + r) / 2;
            for(int i = 0; i < n; i++) {
                double b1 = Y[i] - mid * X[i], c1 = (-b1) / mid;
                double b2 = Y[i] - mmid * X[i], c2 = (-b2) / mmid;
                if(res1 < b1 * c1 * c1) { rr1 = c1; hh1 = b1; res1 = b1 * c1 * c1; }
                if(res2 < b2 * c2 * c2) { rr2 = c2; hh2 = b2; res2 = b2 * c2 * c2; }
            }
            if(res > res1) {
                res = res1; R = rr1; H = hh1;
            }
            if(res > res2) {
                res = res2; R = rr2; H = hh2;
            }
            if(res1 <= res2) r = mmid - eps;
            else l = mid + eps;
        }
        printf("%.3f %.3f\n", H, R);
    }
    return 0;
}