【圆(和球)的最小覆盖】【三分套三分(套三分)】

球的最小覆盖（三分套三分套三分）

 1 /*
 2     三分求球的最小覆盖
 3 */
 4 #include<iostream>
 5 #include<cstdio>
 6 #include<cstring>
 7 #include<algorithm>
 8 #include<cmath>
 9 using namespace std;
10 typedef long long ll;
11 const ll mod=9999973;
12 const double eps=1e-3;
13 int n;
14 double x[105],y[105],z[105];
15 double dis3(double a,double b,double c){
16     double ans=0;
17     for(int i=1;i<=n;i++){
18         ans=max(ans,(x[i]-a)*(x[i]-a)+(y[i]-b)*(y[i]-b)+(z[i]-c)*(z[i]-c));
19     }
20     return ans;
21 }
22 double dis2(double a,double b){
23     double l=-100000;
24     double r=100000;
25     double ans=0;
26     while(r-l>=eps){
27         double rmid=(r+l)/2;
28         double lmid=(l+rmid)/2;
29         if(dis3(a,b,lmid)<dis3(a,b,rmid)){
30             r=rmid;
31         }
32         else l=lmid;
33     }
34     return dis3(a,b,l);
35 }
36 double dis(double a){
37     double l=-100000;
38     double r=100000;
39     while(r-l>=eps){
40         double rmid=(r+l)/2;
41         double lmid=(l+rmid)/2;
42         if(dis2(a,lmid)<dis2(a,rmid)){
43             r=rmid;
44         }
45         else l=lmid;
46     }
47     return dis2(a,l);
48 }
49 int main(){
50    // int n;
51     scanf("%d",&n);
52     for(int i=1;i<=n;i++)scanf("%lf%lf%lf",&x[i],&y[i],&z[i]);
53     double l=-100000;
54     double r=100000;
55     while(r-l>=eps){
56         double rmid=(r+l)/2;
57         double lmid=(l+rmid)/2;
58         if(dis(lmid)<dis(rmid)){
59             r=rmid;
60         }
61         else l=lmid;
62     }
63     printf("%.6lf\n",sqrt(dis(l)));
64     return 0;
65 }
View Code

同理可得圆的最小覆盖（三分套三分）

 

posted @ 2019-03-28 11:45  MekakuCityActor  阅读(245)  评论(0)  编辑  收藏  举报