POJ-2456 Aggressive cows---最大化最小值（也就是求最大值）

题目链接：

https://vjudge.net/problem/POJ-2456

题目大意：

有n个牛栏，选m个放进牛，相当于一条线段上有 n 个点，选取 m 个点， 使得相邻点之间的最小距离值最大 

解题思路：

二分枚举最小距离的最大值

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int INF = 1e9;
const int maxn = 100005;
int n, m;
int a[maxn];
bool judge(int x)
{
    int last = 1;
    for(int i = 1; i < m; i++)
    {
        int now = last + 1;
        while(now <= n && a[now] - a[last] < x)now++;
        last = now;
    }
    return last <= n;
}
int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i++)cin >> a[i];
    sort(a + 1, a + n + 1);
    int l = 0, r = INF + 10, ans;
    while(l <= r)
    {
        int mid = (l + r) / 2;
        if(judge(mid))
            ans = mid, l = mid + 1;
        else r = mid - 1;
    }
    cout<<ans<<endl;
    return 0;
}