- 假設我們枚舉一個半徑r ,那麼分別考慮包含進每個點圓錐的最小高度h0,h1,…,hn−1 ,顯然此時如果半徑爲r ,圓錐高度應爲maxn−1i=0 hi ,簡單推算一下公式,可以發現體積關於半徑是個凹函數,於是三分半徑即可。
- 注意初始的半徑範圍,上限自然可以隨便設置一個較大值,但下限需要稍加考慮。設
di=xi2+yi2−−−−−−−√
那麼r 應滿足r>max{di},0≤i<n
- 輸入輸出部分,即使
ios::sync_with_stdio(false)
也無法避免cin
和cout
的超時問題,所以似乎必須用printf
和scanf
來IO 。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const double EPS = 1e-4;
const double INF = (double)INT_MAX;
const int MAX = 10007;
struct Pos {
double x, y, z;
} a[MAX];
int n;
istream& operator>>(istream& in, Pos& p) {
in >> p.x >> p.y >> p.z;
return in;
}
pair<double, double> gao(double r) {
double h = 0.0;
for (int i = 0; i < n; ++i) {
double dis = a[i].x * a[i].x + a[i].y * a[i].y;
dis = sqrt(dis);
double tp = a[i].z / (r - dis) * dis + a[i].z;
if (tp > h) h = tp;
}
return make_pair(h, r * r * h);
}
pair<double, double> cut(double le, double ri) {
double md, mdmd;
while (ri - le > EPS) {
md = (le + ri) * 0.5;
mdmd = (md + ri) * 0.5;
if (gao(mdmd).second > gao(md).second) {
ri = mdmd;
} else {
le = md;
}
}
return make_pair(gao(le).first, le);
}
int main() {
int T;
scanf(" %d", &T);
while (T--) {
scanf(" %d", &n);
double mindis = 0.0;
for (int i = 0; i < n; ++i) {
scanf(" %lf %lf %lf", &a[i].x, &a[i].y, &a[i].z);
mindis = min(mindis, sqrt(a[i].x * a[i].x + a[i].y * a[i].y));
}
auto res = cut(mindis + EPS, 100000.0);
printf("%.3f %.3f\n", res.first + EPS, res.second + EPS);
}
return 0;
}