二分搜索 POJ 2456 Aggressive cows

题目传送门

/*
    二分搜索：搜索安排最近牛的距离不小于d 
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

const int MAXN = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int x[MAXN];
int n, m;

bool check(int d)   {
    int last = 1;
    for (int i=1; i<=m-1; ++i)    {
        int cur = last + 1;
        while (cur <= n && x[cur] - x[last] < d)    cur++;
        if (cur == n + 1)   return false;
        last = cur;
    }
    return true;
}

int main(void)  {       //POJ 2456 Aggressive cows 
    //freopen ("POJ_2456.in", "r", stdin);

    while (scanf ("%d%d", &n, &m) == 2)   {
        for (int i=1; i<=n; ++i)    {
            scanf ("%d", &x[i]);
        }

        sort (x+1, x+1+n);
        int l = 0, r = INF;
        while (l + 1 < r)   {
            int mid = (l + r) >> 1;
            if (check (mid))    l = mid;
            else    r = mid;
        }
        printf ("%d\n", l);
    }

    return 0;
}