Here you will find solutions of many problems on spoj. If you want solution of some problem which is not listed in blog or have doubt regarding any spoj problem (which i have solved) or any programming concept (data structure) you can mail me @ raj.nishant360@gmail.com

And my humble request to you all that don't copy the code only try to understand the logic and algorithm behind the code. I have started this because if you tried as hard as you can and still can't find any solution to the problem then you can refer to this.
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Saturday, October 17, 2015

WPC5E-Galaxy distances

Galaxy distances

Given below c++ code is for wpc5e spoj or galaxy distances spoj.

Hint :- 
1-> Construct a convex hull from given set of point.(N log N)
2-> Now the points having maximum distance will lie on boundary of convex hull.


/*
===================================================
Name :- Nishant Raj
Email :- raj.nishant360@gmail.com
College :- Indian School of Mines
Branch :- Computer Science and Engineering
Time :- 17 October 2015 (Saturday) 18:16
===================================================*/
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define pll pair < long long , long long >
#define ll long long
#define gc getchar_unlocked
void scanint(int &x)
{
    register int c = gc();
    x = 0;
    int neg = 0;
    for(;((c<48 || c>57) && c != '-');c = gc());
    if(c=='-') {neg=1;c=gc();}
    for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
    if(neg) x=-x;
}
void pr_uint(unsigned long long  n) {
    if (n / 10 != 0)
        pr_uint(n / 10);
    putchar_unlocked((n % 10) + '0');
}

void pr_int(long long n) {
    if (n < 0) {
        putchar_unlocked('-');
        n = -n;
    }
    pr_uint((unsigned long long) n);
}
long long orientation(pll P,pll Q,pll R){
    return  (Q.first-P.first)*(R.second-P.second)-(R.first-P.first)*(Q.second-P.second);
}
int slope_compare(pll &a , pll &b , pll &c , pll &d){
    if((b.second - a.second)*(d.first - c.first) > (b.first - a.first)*(d.second - c.second))
        return 1;
    return 0;
}
void convex_hull(vector<pll > &P,vector<pll > &l,vector<pll > &u){
    int j=0,k=0,n=P.size();
    sort(P.begin(),P.end());
    u.resize(2*n);
    l.resize(2*n);
    for(int i=0;i<n;i++)
    {
        while(j>=2 && orientation(l[j-2],l[j-1],P[i])<=0)
            j--;
        while(k>=2 && orientation(u[k-2],u[k-1],P[i])>=0)
            k--;
        l[j++]=P[i];
        u[k++]=P[i];
    }
    u.resize(k);
    l.resize(j);
}
long long Dist(pll P ,pll Q){
    return abs(P.first*P.first - Q.first*Q.first)+abs(P.second*P.second - Q.second*Q.second);
}
int main()
{
    int t;
    scanint(t);
    while(t--)
    {
        vector<pll >v,u,l;
        int n;
        scanint(n);
        int k1;
        for(int i=0;i<n;i++){
            scanint(k1);
            v.push_back(make_pair(i+1,k1));
        }
        if(n==1){
            printf("0\n");
            continue;
        }
        convex_hull(v,l,u);//conpute convex hull for set of points .
        int i = 0 , j = l.size() -1;
        ll ans = 0;
        l.insert(l.end() , u.begin() , u.end());
        for(int i = 0 ; i < l.size() ; i++){ // Brute Force to find points having maximum distance.
         for(int j = 0 ; j < l.size() ; j++){
          ans = max(ans , Dist(l[i] , l[j]));
         }
        }
        pr_int(ans);
        putchar_unlocked('\n');
    }
}

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