poj 2069

唔。

这道题的火候比较巧妙。

我们是每次找到一个最远的点，然后向那个最远点逼近。

这显然非常合理。

 #include <cstdlib>
 #include <cmath>
 #include <cstdio>
 #include <algorithm>

 using namespace std;
 typedef double db;
 const db eps = 1e-;
 const db delta = 0.98;
 const db INF = 1e100;
 db random(){
     return (rand()&?:-)*rand()*1.0/;
 }
 struct P3{
     db x,y,z;
     P3 operator + (P3 k1){return (P3){x+k1.x,y+k1.y,z+k1.z};}
     P3 operator - (P3 k1){return (P3){x-k1.x,y-k1.y,z-k1.z};}
     P3 operator / (db k1){ return (P3){x/k1,y/k1,z/k1};}
     P3 operator * (db k1){ return (P3){x*k1,y*k1,z*k1};}
     db abs(){return sqrt(x*x+y*y+z*z);}
 };
 P3 p[];
 int n;
 db F(P3 x){
     db res=;
     for(int i=;i<n;i++){
         res=max(res,(x-p[i]).abs());
     }
     return res;
 }
 db SA(){
     db t = ;
     db ans = INF;
     P3 x = {,,};
     while (t>eps){
         db mx = ;int id=-;
         for(int i=;i<n;i++){
             if((x-p[i]).abs()>mx){
                 mx=(x-p[i]).abs();
                 id=i;
             }
         }
         ans = min(ans,mx);
         ans=mx;
         x=x+(p[id]-x)/mx*t;
         t*=delta;
     }
     printf("%.5f\n",ans);
 }
 int main(){
     while (scanf("%d",&n)&&n){
         for(int i=;i<n;i++){
             scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
         }
         SA();
     }
 }