y @ The Angry Teletubbies — Fri, 11 Nov 2011 10:59:00 GMT

【题意】：(x��)在三�l�空间中�Q�给出n个点(x,y,z) (z>=0)�Q�要求用一个体�U�最��的圆锥包围所有的点，输出该圆锥的高和半径�?br />
【题解】：(x��)可以先分析一下，当这个圆锥的高很大时�Q�圆锥的体积非常大；当这个圆锥的高太��的时候，圆锥的半径就�?x��)非常大�Q�此时圆锥的体积�q�是很大�?br /> 于是我们可以得出一个结论，圆锥的高不能太小也不能太大�?br /> �q�里有一步预处理�Q�就是把�I�间直角坐标�p��{换�ؓ(f��)柱面坐标�p�R�?x,y,z) => (sqrt(x*x+y*y),z);
有了上面的结论，我们��可以利用三分极值来��定�q�个高的最优��|��然后通过�怼�三角形就可以求出圆锥的半径�?br /> 此题也可以通过三分半径来求解�?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "cmath"
5 using namespace std;
6 #define maxn 10500
7 #define eps 1e-8
8 #define pi 3.1415926535897932384626433832795
9 int n;
10 double rr;
11 struct Point {
12     double x, y, r, z;
13     Point(){}
14     Point(double _x, double _y, double _z) {
15         x = _x, y = _y, z = _z, r = sqrt(_x * _x + _y * _y) ;
16     }
17 }point[maxn];
18
19 double work(double h) {
20     double tmp;
21     rr = 0.0;
22     for(int i = 0; i < n; i++) {
23         double r = point[i].r, z = point[i].z;
24         double tmp = h * r / (h - z);
25         rr = max(rr, tmp);
26     }
27     double area = pi * rr * rr * h / 3.0;
28     return area;
29 }
30
31 void solve(double res) {
32     double low = res, high = 10000000, mid, mmid, ansh, ansr;
33     while(low + eps < high) {
34         mid = (low + high) / 2.0;
35         mmid = (mid + high) / 2.0;
36         if(work(mid) < work(mmid)) {
37             high = mmid;
38             ansh = mid;
39         } else {
40             low = mid;
41             ansh = mmid;
42         }
43         ansr = rr;
44     }
45     printf("%.3f %.3f\n", ansh, ansr);
46
47 }
48
49 int main() {
50     int T;
51     double x, y, z;
52     scanf("%d", &T);
53     while(T--) {
54         scanf("%d", &n);
55         double low = 0.0;
56         for(int i = 0; i < n; i++) {
57             scanf("%lf%lf%lf", &x, &y, &z);
58             point[i] = Point(x, y, z);
59             low = max(low, z);
60         }
61         solve(low);
62     }
63     return 0;
64 }

y @ The Angry Teletubbies 2011-11-11 18:59 发表评论

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