F - Aggressive cows

最新推荐文章于 2025-07-11 17:47:48 发布

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解决农场中牛棚布局问题，通过二分查找算法找到最大的最小距离以避免牛之间的争斗。输入包括牛棚数量和牛的数量，以及各牛棚的位置。

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.

大佬程序设计的太巧妙，二分竟然可以这样使用，自愧不如，附上代码一枚时常品味

#include<iostream>
#include<algorithm>

using namespace std;

typedef long long ll;

const long maxn=1e5+5;

ll x[maxn];
long n,c;

bool f(ll m)
{
    ll last=0,cur;
    for(int i=1;i<c;i++)
    {
        cur=last+1;
        while(cur<n&&x[cur]-x[last]<m)cur++;
        if(cur==n)return false;
        last=cur;
    }
    return true;
}

int main()
{
    cin>>n>>c;
    for(long i=0;i<n;i++)
        cin>>x[i];
    sort(x,x+n);
    ll l=x[0],r=x[n-1],mid;
    while(r-l>1)
    {
        mid=(l+r)/2;
        if(f(mid))l=mid;
        else r=mid;
    }
    cout<<l<<endl;
    return 0;
}