Third (minimization of the maximum value)

https://ac.nowcoder.com/acm/contest/3006/B

The meaning of problems: there are n training base, the coordinates xi, yi (-10000 <= x, y <= 10000), the x-axis to build a maximum value of minimum venue Shidao all training base. Find the value.

Solution: The answer is a clear function of single valley, close to one-third the answer.

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define cin(x) scanf("%lld" , &x);
using namespace std;
const int N = 1e7+9;
const int maxn = 1e5+9;
const double esp = 1e-6;
pii a[maxn];
int n ;

double cal(double x){
    double ans = -INF;
    rep(i , 1 , n){
        ans = max(ans , sqrt((a[i].fi-x)*(a[i].fi-x) + a[i].se*a[i].se));
    }
    return ans ;
}

void solve(){
    cin >> n ;
    rep(i , 1 , n){
        cin >> a[i].fi >> a[i].se;
    }
    double l = -1e4 , r = 1e4 ;
    while(r - l >= esp){
        double rmid = r - (r - l) / 3 , lmid = l + (r - l) / 3 ;
        if(cal(rmid) >= cal(lmid)){
            r = rmid ;
        }else{
            l = lmid;
        }
    }
    printf("%.4lf\n" , cal(r));
}

signed main()
{
    ios::sync_with_stdio(false);
    //cin.tie(0); cout.tie(0);
    //int t ;
    //cin(t);
    //while(t--){
        solve();
    //}
}