Problem D. Country Meow 题解(三分套三分套三分)

题目链接

题目大意

给你n(n<=100)个点，要你找一个点使得和所有点距离的最大值最小值ans

题目思路

一直在想二分答案，但是不会check

这个时候就要换一下思想

三分套三分套三分坐标即可

复杂度\(O(n(log_n)^3)\)

代码

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#define fi first
#define se second
#define debug printf(" I am here\n");
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e2+5,inf=0x3f3f3f3f,mod=1e9+7;
const double eps=1e-5;
int n;
double  x[maxn],y[maxn],z[maxn];
double check3(double a,double b,double c){
    double ma=-1;
    for(int i=1;i<=n;i++){
        ma=max(ma,sqrt((a-x[i])*(a-x[i])+(b-y[i])*(b-y[i])+(c-z[i])*(c-z[i])));
    }
    return ma;
}
double check2(double x,double y){
    double zl=-1e5,zr=1e5;
    while(zl+eps<=zr){
        double zmid1=(zr-zl)/3+zl,zmid2=2*(zr-zl)/3+zl;
        if(check3(x,y,zmid1)<check3(x,y,zmid2)){
            zr=zmid2;
        }else{
            zl=zmid1;
        }
    }
    return check3(x,y,zl);
}
double check1(double x){
    double yl=-1e5,yr=1e5;
    while(yl+eps<=yr){
        double ymid1=(yr-yl)/3+yl,ymid2=2*(yr-yl)/3+yl;
        if(check2(x,ymid1)<check2(x,ymid2)){
            yr=ymid2;
        }else{
            yl=ymid1;
        }
    }
    return check2(x,yl);
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lf%lf%lf",&x[i],&y[i],&z[i]);
    }
    double xl=-1e5,xr=1e5;
    while(xl+eps<=xr){
        double xmid1=(xr-xl)/3+xl,xmid2=2*(xr-xl)/3+xl;
        if(check1(xmid1)<check1(xmid2)){
            xr=xmid2;
        }else{
            xl=xmid1;
        }
    }
    printf("%.10f\n",check1(xl));
    return 0;
}