最大化最小值poj2456Aggressive cows

Aggressive cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15528		Accepted: 7440

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.

check（d）=可以安排所有的牛的位置使得最近的两头牛的距离都不小于d

求满足check（d）的最大的d的值。

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
int house[maxn];
int n,m;
const int inf=99999999;
bool check(int d)
{
    int last=0;
    for(int i=1;i<m;i++)
    {
        int crt=last+1;
        while(crt<n&&house[crt]-house[last]<d)
        {
            crt++;
        }
        if(crt==n)return false;
        last=crt;
    }
    return true;
}
void solve()
{
    sort(house,house+n);
    int lb=0,ub=inf;
    while(ub-lb>1)
    {
        int mid=(ub+lb)/2;
        if(check(mid))
            lb=mid;
        else
            ub=mid;
    }
    printf("%d
",lb);
}

int main()
{
    cin>>n>>m;
    for(int j=0;j<n;j++)
        scanf("%d",&house[j]);
    solve();
    return 0;
}