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.


hide comments
garry_555: 2019-07-28 08:20:42

nyc qs on binary search! must do!

anshika_85: 2019-07-04 12:48:43

AC in one go.!!

dbzadnen: 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
anshuman1117: 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
alok_0366: 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??

abhay1999: 2019-05-22 22:05:45

use basic knowledge of Binary search and sorting..AC in one go

ansu2678: 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
phuongbo_255: 2019-05-10 17:30:00

use binary search to find the answer

harshraj22aug: 2019-05-07 23:30:43

quora explaination ( https://qr.ae/TWIaVc ) was way too better than topcoder tutorial. One of the very good questions of binary search that I have seen. If anyone gets similar question, please comment here .

sandeep48: 2019-03-26 18:46:53

AC in one go..easy implementation of binary search


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