【POJ 2456】 （二分法 &&）

Aggressive cows

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 16433 Accepted: 7865

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.

关于最大化最小值或者最小化最大值的问题，通常用二分搜索法。

#include<cstdio>
#include<algorithm>
using namespace std;
int a[100005],m,n;
int check(int len)
{
      int x,sum=1;
      x=a[1];
      for(int i=2; i<=n; i++)
      {
      if(a[i]-x>=len)
      {
      ++sum;
      x=a[i];
 }
 }
 if(sum>=m)
 {
  return 1;
 }
 else
 return 0;
 } 
 int main()
 {
  int ans=0;
  scanf("%d%d",&n,&m);
  for(int i=1; i<=n; i++)
  scanf("%d",&a[i]);
  sort(a+1,a+n+1);
int L=0,R=a[n]-a[1],mid;
    while(R>=L)
    {
    mid=(R+L)/2;
    if(check(mid)==1)
    {
    ans=mid;
    L=mid+1;
}
else
R=mid-1;
}
printf("%d\n",ans);
return 0;
 
 }