POJ2420

三分和二分输出答案是l还是r就看中间的判断条件，不满足条件更新的那个值就是要输出的值

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#include <iostream>
#include <cmath>
#include <iomanip>
#define ios ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define endl '\n'
using namespace std;
const int N=1e3+100;
const double eps=1e-5;
int n;
int a[N],b[N];
double getdis(double x,double y){
	double res=0;
	for(int i=1;i<=n;i++) res+=sqrt(pow(x-a[i],2)+pow(y-b[i],2));
	return res;
}
double work(double x){
	double l=0,r=10000,lmid,rmid;
	while(r-l>eps){
		lmid=l+(r-l)/3;
		rmid=r-(r-l)/3;
		if(getdis(x,lmid)>getdis(x,rmid)) l=lmid;
		else r=rmid;
	}
	return getdis(x,lmid);
}
int main()
{
	ios;
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i]>>b[i];
	double l=0,r=10000,lmid,rmid;
	while(r-l>eps){
		lmid=l+(r-l)/3;
		rmid=r-(r-l)/3;
		if(work(lmid)>work(rmid)) l=lmid;
		else r=rmid;
	}
	cout<<fixed<<setprecision(0)<<work(l)<<endl;
	return 0;
}