三分套三分再套三分

最新推荐文章于 2021-07-23 14:43:51 发布

原创最新推荐文章于 2021-07-23 14:43:51 发布 · 354 阅读

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南京赛的最小球覆盖，求最小的球的半径南京赛最小球覆盖

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
const int maxn = 100+20;
const double eps = 1e-3;
using namespace std;

int n;
struct node{
    double x;
    double y;
    double z;
}e[maxn];

double f3(double x,double y,double z){
    double ans = 0;
    for(int i = 0 ; i < n ; i++)
        ans = max(ans,(e[i].x-x)*(e[i].x-x) + (e[i].y-y)*(e[i].y-y) + (e[i].z-z)*(e[i].z-z));
    return ans;
}

double f2(double x,double y){
    double l = -100000.0;
    double r = 100000.0;
    while(r - l > eps){
        double lmid = l + (r-l)/3.0;
        double rmid = r - (r-l)/3.0;
        if(f3(x,y,lmid) < f3(x,y,rmid)) r = rmid;
        else l = lmid;
    }
    return f3(x,y,l);
}

double f1(double x){
    double l = -100000.0;
    double r = 100000.0;
    while(r - l > eps){
        double lmid = l + (r-l)/3.0;
        double rmid = r - (r-l)/3.0;
        if(f2(x,lmid) < f2(x,rmid)) r = rmid;
        else l = lmid;
    }
    return f2(x,l);
}

int main(){
    while(~scanf("%d",&n)){
        for(int i = 0 ; i < n ; i++){
            scanf("%lf%lf%lf",&e[i].x,&e[i].y,&e[i].z);
        }
        double l = -100000.0, r = 100000.0;
        while(r - l > eps){
            double lmid = l + (r-l)/3.0;
            double rmid = r - (r-l)/3.0;
            if(f1(lmid) < f1(rmid)) r = rmid;
            else l = lmid;
        }
        printf("%.5f\n",sqrt(f1(l)));
    }
    return 0;
}