HackerRank Fundamentals – Most Distant

Challenge: Most Distant

Subdomeniu: Fundamentals (fundamentals)

Scor cont: 30.0 / 30

Submission status: Accepted

Submission score: 1.0

Submission ID: 464726522

Limbaj: cpp14

Link challenge: https://www.hackerrank.com/challenges/most-distant/problem

Cerinta

Keko has $N$ dots in a 2-D coordinate plane. He wants to measure the gap between the most distant two dots. To make the problem easier, Keko decided to change each dot's $x$ *or* $y$ coordinate to zero.   

Help Keko calculate the distance!

Input Format

The first line contains an integer, $N$, the number of dots.  
The next $N$ lines each contain the integer coordinates of the dots in $(x, y)$ fashion.  

**Constraints**  
$2\leq N \leq 10^6$  
$-10^9\leq x_i,\,y_i \leq 10^9$

It is guaranteed that all dots are distinct, and either their $x$ or $y$ coordinate is equal to $0$.

Output Format

Print the distance between the most distant dots with an absolute error of, at most, $10^{-6}$.

Cod sursa

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    double mnx,mxx,mny,mxy; //min_x_axis  max_x_axis  min_y_axis  max_y_axis  
    int t;
    cin>>t;
    while(t--)
    {
        double a,b;
        cin>>a>>b;
        if(a==0)
        {
            mny=min(mny,b);
            mxy=max(mxy,b);
        }
        else
        {
            mnx=min(mnx,a);
            mxx=max(mxx,a);
            
        }
        
        
    }
        double mx=max(abs(mxx-mnx),abs(mny-mxy));//axis
        double q1=sqrt(mxx*mxx+mxy*mxy); //quadrant 1
        double q2=sqrt(mnx*mnx+mxy*mxy);    //2
        double q3=sqrt(mnx*mnx+mny*mny);    //3
        double q4=sqrt(mxx*mxx+mny*mny);    //4
    
        printf("%.10f",max(mx,max(q1,max(q2,max(q3,q4)))));
    
    return 0;
}