POJ2456Aggressive cows-（二分判定）

Posted 2021-11-08 shoulinniao

Description

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
翻译：有n间牛房在一条直线上，距离分别是xi，有c只牛，牛在一起容易打架，让牛尽可能在距离比较远的牛房。要安置c只牛，求这个距离最大能是多少？
白书上的二分题目，书上讲得一点都不人性化，能用人话说清楚的非要用鬼话，代码也是不够简洁，看了半个小时都不懂，一搜题解，秒懂，记录一下代码。

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<math.h>
 6 #include<string>
 7 #define ll long long
 8 #define inf 0x3f3f3f3f
 9 using namespace std;
10 int n,m;
11 int a[100086];
12 
13 bool check(int d)
14 {
15     int num=1,now=a[0];
16     for(int i=1;i<n;i++)
17     {
18         if(a[i]-now>=d)
19             now=a[i],num++;///如果a[i]号牛棚距离上一个的距离>=d,更新牛棚位置，并且增加安置的牛数，大于等于m时候就是可行解
20         if(num>=m)
21             return true;
22     }
23     return false;
24 }
25 
26 int main()
27 {
28     while(scanf("%d %d",&n,&m)!=EOF)
29     {
30         for(int i=0;i<n;i++)
31             scanf("%d",&a[i]);
32         sort(a,a+n);
33         int l=1,r=a[n-1],ans=0;
34         while(l<=r)
35         {
36             int mid=(l+r)/2;
37             if(check(mid))
38             {
39                 ans=mid;///不断更新可行解，找到最大值， 即最大化 可行解。
40                 l=mid+1;
41             }
42             else
43                 r=mid-1;
44         }
45         printf("%d\n",ans);
46     }
47     return 0;
48 }