POJ-2456- Aggressive cows

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. 

 

1. 从1到最远的距离二分枚举距离。判断是否满足条件。

2. 判断过程中，距离不够就继续加。够了就计数加一。最后判断计数和牛的数量。如果计数少了。那么距离大了。如果计数多了。距离太小。

3. 根绝返回值的情况。必须保证计数大于等于牛的数量，也就是最后符合的值是放在了l-1上。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int a[100002];
int n,c;
bool check(int mid)
{
    int sum = 0,cnt = 1;
    for(int i=n-1;i>=1;i--)
    {
        if(a[i+1]-a[i]+sum>=mid)
        {
            cnt++;
            sum = 0;
        }
        else
        {
            sum+=a[i+1]-a[i];
        }
    }
    if(cnt<c)
        return false;
    else
        return true;
}
int main()
{
    while(~scanf("%d%d",&n,&c))
    {
        int l=1,r = -1,mid;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a+1,a+n+1);
        r = a[n];
        while(r>=l)
        {
            mid = (l+r)>>1;
            if(check(mid))
                l = mid+1;
            else
                r = mid-1;
        }
        printf("%d\n",l-1);
    }
    return 0;
}