AGGRCOW - Aggressive cows

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.


Added by:Roman Sol
Date:2005-02-16
Time limit:2s
Source limit:10000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All
Resource:USACO February 2005 Gold Division

hide comments
2019-09-02 17:08:23
great ques but poorly worded
2019-08-25 12:10:34
My code working on my computer but showing wrong answer in SPOJ

#include<bits/stdc++.h>
using namespace std;
int barn, cow, temp;
vector<int> pos;

bool place_them(int mid)
{
int count=1, start=pos[0];
for(int i=0; i<barn; i++)
{
if(pos[i]-start >= mid)
{
count++;
start = pos[i];
}
if(count==cow) return 1;
}
return 0;
}

int binary(int lo, int hi)
{
int mid;
while(hi-lo>1)
{
mid = lo + (hi-lo)/2;
if(place_them(mid)) lo = mid;
else hi = mid;
}
return mid;
}

int main()
{
int t;
cin >> t;
while(t--)
{
cin >> barn >> cow;
pos.clear();

for(int i=0; i<barn; i++)
{
cin >> temp;
pos.push_back(temp);
}
sort(pos.begin(), pos.end());
int lo, hi;
lo = 0;
hi = pos[barn-1] - pos[0] + 1;
temp = binary(lo, hi);
cout << temp << endl;
}
return 0;
}


Last edit: 2019-08-25 12:11:09
2019-08-17 10:58:01
great question on binary search!!
2019-07-28 08:20:42
nyc qs on binary search! must do!
2019-07-04 12:48:43
AC in one go.!!
2019-06-30 17:15:45
can someonne highlight my error , it's a wrong answer while the output is 3
#include <iostream>
#include<vector>
#include<algorithm>

using namespace std;

int check(int i, vector<int> vec,int c) {
int count = 1;
int pos = 0;
int last = 0;
int maxDistance = vec.back() - vec.front();
while (pos < (vec.size())) {
if (vec[pos] - vec[last] >= i) {
count++;
maxDistance = min(maxDistance, vec[pos] - vec[last]);
last = pos;
}
pos++;
if (count == c)
return maxDistance;
}
return 0;
}

int main() {
int t;
cin >> t;
for (int i = 0; i < t; i++) {
int pos, c;
cin >> pos >> c;
vector<int> vec;
for (int j = 0; j < pos; j++) {
int p;
cin >> p;
vec.push_back(p);
}
sort(vec.begin(), vec.end());
int start = vec[0];
int end = vec.back();
int score = 0;
while (start < end) {
int mid = start + (end - start) / 2;
score = check(mid, vec, c);
if (score>0) {
start = mid + 1;
}
else {
end = mid -1;
}
}
cout << score << endl;
}
}

Last edit: 2019-06-30 17:16:12
2019-06-25 07:30:56
Those who are not getting it, check the first question solved in this video, from 5:10. : https://youtu.be/TC6snf6KPdE

Last edit: 2019-06-25 07:32:01
2019-06-16 10:27:28
is it ncecessary that we have to implement it only under the tag under which the ques is categorised..that is DP??
2019-05-22 22:05:45
use basic knowledge of Binary search and sorting..AC in one go
2019-05-18 08:11:45
sort(all(a));
ll low = 0;
ll high = a[n-1]-a[0]+1;
ll mid;
while(low < high)
{
//cout<<low<<" "<<high<<endl;
mid = (low+high+1)/2;
if(check(mid,a,c))
low = mid;
else
high = mid-1;
}
cout<<low;



what is problem with this code plz tell

Last edit: 2019-05-18 08:12:14
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