poj 2456 Aggressive cows （二分+贪心思想）

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. 

题意：

FJ 有一个很长的 barn ，里面有 N 棚， 假设N 个棚在一条直线上，第 i 个棚的位置为 xi.

然后他有 c 只羊，为了防止羊相互攻击，要找出最大的两只羊之间距离，前提是这 c 只羊都必须能放下。

解题思路：先将 xi 由小到大排序， xn 最大。设这个最大距离为 s;0=<s<=(xn-x1), 因为 c 只羊需要 c-1 个间隔。使用二分找 s 的可能取值

代码如下：

#include <iostream>
#include <algorithm>
using namespace std;
int a[100010];
int main()
{
	int n,m;
	while(cin>>n>>m)
	{
		int high,low,mid;
		for(int i=0;i<n;i++)  cin>>a[i];
		sort(a,a+n);  // 必须先排序 
		high=a[n-1]-a[0],low=0; //low的值必须考虑0 
		int left=0,right=high,cnt;
		for(int i=0;i<100;i++)
		//while(left+1<right)  while和for都可 
		{
			cnt=1;
			mid=(left+right)/2;
			int temp=a[0];
			for(int i=1;i<n;i++)
			{
				if((a[i]-temp)>=mid)
				{
					cnt++;
					temp=a[i];
				}
			}
			if(cnt>=m) left=mid;
			else right=mid;
		}
		printf("%d\n",left);
	 } 
	 return 0;
 } 

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