UVA 1473 - Dome of Circus(三分+计算几何)

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这题先推下公式，很容易发现圆锥的体积和半径满足一个二次函数，所以可以用三分

于是三分半径即可

代码：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int N = 10005;const double PI = acos(-1.0);struct Point {    double x, y, z;    void read() {        scanf("%lf%lf%lf", &x, &y, &z);    }} p[N];int n;double gao(double r) {    double h = 0;    for (int i = 0; i < n; i++) {        double d = sqrt(p[i].x * p[i].x + p[i].y * p[i].y);        h = max(h, p[i].z / (r - d) * r);    }    return h;}int main() {    while (~scanf("%d", &n)) {        double l = 0, r = 1e20;        for (int i = 0; i < n; i++) {            p[i].read();            l = max(l, sqrt(p[i].x * p[i].x + p[i].y * p[i].y));        }        l += 1e-8;        for (int i = 0; i < 200; i++) {            double midl = (l * 2 + r) / 3;            double midr = (l + r * 2) / 3;            double h1 = gao(midl);            double h2 = gao(midr);            double s1 = PI * midl / 3 * midl * h1;            double s2 = PI * midr / 3 * midr * h2;            if (s1 > s2) l = midl;            else r = midr;        }        printf("%.3f %.3f\n", gao(l), l);    }    return 0;}

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