HDU ACM 2456 Aggressive cows

最新推荐文章于 2019-09-25 19:35:49 发布

转载最新推荐文章于 2019-09-25 19:35:49 发布 · 151 阅读

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/i8888/p/4044033.html

FarmerJohn建造了一个新长栏，拥有N个栏位，位于直线上。C头牛不喜欢这种布局，彼此间一旦进入同一栏就会变得攻击性。FJ希望通过合理的分配牛到栏位，使得两头牛之间的最小距离尽可能大。此博客提供了解决该问题的算法，包括输入格式、输出格式及示例输入输出。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

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.

代码如下：

#include<iostream>
#include<algorithm>
using namespace std;


long N,C,a[100010];


bool search(long x)
{
long i,n=a[1],m=1;
for (i=2;i<=N;i++)
{
if (a[i]-n>=x)
{
   n=a[i];
m++;
}
if (m>=C)
   return true;
}
return false;
}


int main()
{
while(cin>>N>>C)
{
for(int i=1; i<=N; i++)  
            cin>>a[i];  
        sort(a+1,a+1+N);  //   高效的排序方法（升序）
        int l=0,r=a[N],mid;  
        while(l+1<r)  
        {  
            mid=(l+r)/2;  
            if(search(mid))  
                l=mid;  
            else  
                r=mid;  
        }  
        cout<<l<<endl;  
}
return 0;
}

转载于:https://www.cnblogs.com/i8888/p/4044033.html