ACM-oj -G

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 312849

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.

题意：x轴上有N个点，你要选出C个点来，使得他们之间最近的距离最远。

方法：二分法

代码：

  #include<iostream>  #include<algorithm>  #include<cmath>  //#define ll long long  using namespace std;    int a[100005];  int n,c;    bool f(int mid)  {  int s = 0,l = 1;  for(int i = 1;i < n; ++i){  if((s + a[i] -a[i-1]) < mid)  {  s += a[i]-a[i-1];}else{s = 0;++l;}} if(l < c){return false;}else{return true;}  }   int main()  {  cin >> n >> c;  for(int i=0;i<n;++i)  {  cin >> a[i];}sort(a,a+n);int low = a[n-1] - a[0];int high = a[n-1] - a[0];                      //找出最大距离for(int i = 1;i < n; ++i){low = min(low,a[i]-a[i-1]);                //找出最小距离}while(low <= high)                             //将所有距离使用二分法，找出最优解{int mid = (low+high)/2;                    //假设mid是最优距离           if(f(mid))                                 //数组中距离大于mid的个数大于c，说明mid值偏小{low = mid + 1;                        }else                                       //mid值偏大{high = mid - 1;                       }}cout << high << endl;  return 0;  }
要点：
要先把坐标排序，然后用在最短距离跟最大距离之间进行查找，每次二分判断最小距离为mid的个数.