POJ - 2456 Aggressive cows

Posted 2020-10-16 小小超plus

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 want 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

* Line 1: Two space-separated integers: N and C 

* Lines 2..N+1: Line i+1 contains an integer stall location, xi

Output

* Line 1: One integer: the largest minimum distance

Sample Input

5 3
1
2
8
4
9

Sample Output

3

Hint

OUTPUT DETAILS: 

FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3. 

Huge input data,scanf is recommended.

 1 #include <iostream>
 2 using namespace std;
 3 #include<string.h>
 4 #include<set>
 5 #include<stdio.h>
 6 #include<math.h>
 7 #include<queue>
 8 #include<map>
 9 #include<algorithm>
10 #include<cstdio>
11 #include<cmath>
12 #include<cstring>
13 #include <cstdio>
14 #include <cstdlib>
15 #include<stack>
16 #include<vector>
17 long long n,k;
18 long long a[110000];
19 long long panduan(long long zhi)
20 {
21     int add=1;
22     int kaishi=1,jieshu=1;
23     while(1)
24     {
25         if(a[jieshu]-a[kaishi]<zhi&&jieshu<=n)
26         {
27             jieshu++;
28         }
29         else if(a[jieshu]-a[kaishi]>=zhi)
30         {
31             add++;
32             kaishi=jieshu;
33         }
34         else
35             break;
36     }
37     //cout<<zhi<<"_"<<add<<endl;
38     return add;
39 }
40 int  main()
41 {
42     scanf("%lld %lld",&n,&k);
43     for(long long  i=1;i<=n;i++)
44         scanf("%lld",&a[i]);
45     sort(a+1,a+1+n);
46     long long kaishi=0,jieshu=1e18;
47     long long mid;
48     long long t;
49     while(kaishi<jieshu-1)
50     {
51         //cout<<kaishi<<"_"<<jieshu<<endl;
52         mid=(kaishi+jieshu)>>1;
53         if(panduan(mid)>=k)
54             kaishi=mid;
55         else
56             jieshu=mid;
57     }
58     cout<<kaishi<<endl;
59     return 0;
60 }

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