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最新推荐文章于 2020-09-16 16:11:14 发布

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本文介绍了一个经典的算法问题：如何在给定的牛棚位置中放置一定数量的牛，使得任意两头牛之间的最小距离最大化。文章通过二分查找的方法解决该问题，并提供了完整的C++实现代码。

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.

题目大意是让牛之间的间距最大化，即最大化最小值（同样最小化最大值也用二分）。

#include <cstdio>
#include <cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
ll a[100005];
int n,c;
bool judge(ll x)
{
   int sum=1;
   ll k=a[0];
   for(int i=1;i<n;i++){
    if(a[i]-k>=x)
    {
        k=a[i];
        sum++;
        if(sum>=c)
            return true;
    }
   }
   return false;
}
void solve()
{
    ll right=a[n-1],left=0,mid;
    while(right>=left){//二分法要注意1.循化出口的条件，否则会死循环 2.选择左边好是右边的问题；
        mid=(left+right)/2;
        if(judge(mid))
            left=mid+1;
        else
            right=mid-1;
    }
    cout<<left-1<<endl;
}
int main()
{
    scanf("%d %d",&n,&c);
      for(int i=0;i<n;i++)
        scanf("%ld",&a[i]);//一定要用scanf，否则100000的数量级会超时
            sort(a,a+n);//贪心要想到排序。
        solve();
   
    return 0;
}