Editorial for Aggressive Cows-AGGRCOW problem SPOJ - UCS - Unleash-Coding-Skills

Sunday, 4 March 2018

Editorial for Aggressive Cows-AGGRCOW problem SPOJ - UCS

this is solution for Aggressive cows problem in spoj here we used binary search in a different way..
Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).

His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ wants to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

Input

t – the number of test cases, then t test cases follows.
* Line 1: Two space-separated integers: N and C
* Lines 2..N+1: Line i+1 contains an integer stall location, xi

Output

For each test case output one integer: the largest minimum distance.

Example

Input:
1
5 3
1
2
8
4
9
Output:
3
Output details:
FJ can put his 3 cows in the stalls at positions 1, 4 and 8,
resulting in a minimum distance of 3.
#include <iostream>

#include <algorithm>

#include <vector>

using namespace std;

vector<long long int>v;

long long int n,C;

long long int fun(long long int mid)

{

  long long   int cows=1,pos=v[0];    



 for(long long int i=1;i<n;i++)

    {

        if(v[i]-pos>=mid)

        {

            pos=v[i];

            cows++;



            if(cows==C)

                return 1;



        }

    }

    return 0;

}







int main() {



long long int i,j,k,mid,o,fo,max_dis,lo,hi,uv,t;

cin>>t;

for(i=0;i<t;i++)

{

      cin>>n>>C;

         v.clear();

      for(j=0;j<n;j++)

      {

      cin>>k;



      v.push_back(k);



      }

      sort(v.begin(),v.end());

      o=0;

      max_dis=1;

      lo=0;

      hi=v[n-1];

      while(lo<hi)

      {

     mid=lo+(hi-lo)/2;

            if(fun(mid))

            {

                  if(max_dis<mid)

                  {

                        max_dis=mid; 



                  }

             lo=mid+1;

            }

            else

            {

                  hi=mid;

            }

}

    cout<<max_dis<<endl;   

}

 return 0;

}

1 comment:

  1. It would have been better if you explained the solution as well along with the code.

    ReplyDelete