E - Aggressive cows

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.

把C头牛放到N个带有编号的隔间里，使得任意两头牛所在的隔间编号的最小差值最大。例如样例排完序后变成1 2 4 8 9，那么1位置放一头牛，4位置放一头牛，它们的差值为3；最后一头牛放在8或9位置都可以，和4位置的差值分别为4、5，和1位置的差值分别为7和8，不比3小，所以最大的最小值为3。

分析：这是一个最小值最大化的问题。先对隔间编号从小到大排序，则最大距离不会超过两端的两头牛之间的差值，最小值为0。所以我们可以通过二分枚举最小值来求。假设当前的最小值为x，如果判断出最小差值为x时可以放下C头牛，就先让x变大再判断；如果放不下，说明当前的x太大了，就先让x变小然后再进行判断。直到求出一个最大的x就是最终的答案。

#include<iostream>
#include<algorithm>
#include<cstdio>
#define P 1000000000
using namespace std;
int x[100010];
int N,M;
bool ad(int t)t为两头牛的距离
{
    int last=0;
    for(int i=1;i<M;i++){
        int cur=last+1;
        while(cur<N&&x[cur]-x[last]<t)//判定cur位置是否能够放牛，如果能，不会进入循环
            cur++;
        if(cur==N)return false;//当for循环还没结束，cur==N时说明到最后一个位置是所有的牛还没有安排完，此时返回false
        last=cur;//把第i头牛放在编号为cur的牛舍中，这里牛的数量和编号都是从0开始
    }
    return true;//能在cur到达最后位置之前跳出for循环说明能够安排所有的牛，所以返回true
}
int main()
{
    //freopen("1.txt","r",stdin);
    while(cin>>N>>M)
    {
        for(int i=0;i<N;i++)cin>>x[i];
        sort(x,x+N);
        int low=0,high=P;
        while(high-low>1){
        int mid=(high+low)>>1;
        if(ad(mid))low=mid;
        else high=mid;
        }
        printf("%d\n",low);

    }

    return 0；
}