二分贪心 E

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 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.

           这道题的基本题意为给你n个畜栏、c头牛，把这c头牛放在其中c个畜栏里，让牛与牛之间离得尽可能的远，要求求满足条件的牛与牛之间的最小距离。

源代码如下：

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
int main()
{ int n,c,a[100001],i,j,mid,min,max;
  scanf("%d%d",&n,&c);
  for(i=0;i<n;++i)
   scanf("%d",&a[i]);
  sort(a,a+n);
   min=0;
   max=a[n-1];
   while(max-min>1)
   { int m=1;
     mid=(min+max)/2;
     for(i=1,j=0;i<n;++i)
      if(a[i]-a[j]>=mid)
      { ++m;
        j=i;
 }
 if(m>=c)
   min=mid;
 else max=mid;
   }
    printf("%d\n",min);
 } 

         需要注意的是while里的循环条件要用max-min>1，如果用max>min的话，则无限循环没有结果输出。