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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
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2019-12-24 15:20:19
Poor Explanation |
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2019-12-17 20:39:46
AC in one go !!! good question on BinarySearch !!! must do !! |
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2019-12-08 10:10:49
interesting problem |
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2019-10-26 08:52:01
AC in one go! Good question but sample test case could've been better. |
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2019-10-17 18:34:03
Test cases are not up to mark showing correct answer even if coded wrong |
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2019-09-29 13:22:29
Finally One Correct in One Go!! |
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2019-09-28 15:29:56
Last edit: 2019-09-28 15:34:15 |
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2019-09-12 00:27:11
Really good question about binary search. |
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2019-09-02 17:08:23
great ques but poorly worded |
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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 |
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