Aggressive cows（二分）

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农夫约翰新建了一条长廊式牛棚，包含N个摊位。为了防止C头牛因相邻而变得好斗，他要找出最大的最小摊位间距以避免冲突。输入包括牛棚摊位位置和牛的数量，输出应为能实现的最大最小距离。样例输入显示在5个摊位中安排3头牛，最大最小距离可达到3。

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
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.

n个房子，c只牛，每个牛都不想距离太近，问牛之间能达到的最小距离。

直接二分房子之间距离，当前这只牛在x，那么x + mid之间的房子都不能有牛，看牛是否可以分配完，直接用lower_bound就可以。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<utility>
#include<sstream>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int n,c;
ll a[100005];
bool ok(int x)
{
    int pre = 1;
    for(int i = 2;i <= c;i++)
    {
        pre = lower_bound(a + pre,a + 1 + n,a[pre] + x) - a;
        if(pre > n)return false;
    }
    return true;
}
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\巍巍\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\巍巍\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    scanf("%d%d",&n,&c);
    for(int i = 1;i <= n;i++)
        scanf("%lld",&a[i]);
    sort(a + 1,a + 1 + n);
    int l = 0,r = inf;
    while(r - l > 1)
    {
        int mid = (r + l)/2;
        if(ok(mid))
            l = mid;
        else
            r = mid;
    }
    printf("%d\n",l);
    return 0;
}