Aggressive cows（二分）

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;}