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-7;
const db delta = 0.98;
const db INF = 1e100;
db random(){
    return (rand()&1?1:-1)*rand()*1.0/32767;
}
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[33];
int n;
db F(P3 x){
    db res=0;
    for(int i=0;i<n;i++){
        res=max(res,(x-p[i]).abs());
    }
    return res;
}
db SA(){
    db t = 100;
    db ans = INF;
    P3 x = {0,0,0};
    while (t>eps){
        db mx = 0;int id=-1;
        for(int i=0;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
",ans);
}
int main(){
    while (scanf("%d",&n)&&n){
        for(int i=0;i<n;i++){
            scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
        }
        SA();
    }
}