POJ2456——二分——Aggressive cows

/*
二分经典题
大意  n头牛让你放m头要求这些牛之间的最小距离在所有情况里面的最大值
判断的时候移动下标
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int inf = 0x3f3f3f3f;
const int MAX = 1e6 + 10;
int n, m;
int a[MAX];

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

int main()
{
    while(~scanf("%d%d", &n, &m)){
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        sort(a + 1, a + n + 1);
        int l = 0, r = inf;
        while(l <= r){
            int mid = (l + r) >> 1;
            if(check(mid)){
            l = mid + 1;
            }
            else r = mid - 1;
        }
        printf("%d\n", l);
    }
    return 0;
}