POJ 2456 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.

题意：农夫有c头牛，n个隔间，c头牛很躁动，很容易相互打架，因此农夫想把它们分得越远越好，要你分配隔间使得相邻两头牛的距离越远越好，问你这c头牛分割的最小距离的最大值。

思路：二分的思想，最小距离为1，最大距离为牛栏编号最大的减去编号最小的 

AC代码:

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,c;
long long a[100005];
int judge(int x)
{
	long long index=a[0];
	int num=1;  //当前一头已经放进了a[0] 
	for(int i=0;i<n;i++)
	{
		if(a[i]-index>=x) 
		{
			num++;
			index=a[i];
		}
		if(num>=c) return 1; 	 
	}
	return 0;
}
void solve()
{
	int l = 1,r = a[n-1]-a[0];

	while(l<=r)
	{
		int mid=(l+r)>>1;
			//cout<<mid<<endl;
		if(judge(mid)) l=mid+1;
		else r=mid-1;
	
	}
	printf("%d\n",r);
	
}
int main()
{
	scanf("%d%d",&n,&c);
	for(int i=0;i<n;i++)
	scanf("%lld",&a[i]);
	sort(a,a+n);
	solve();
	return 0;
}