本题还有一个低配版：1049. 大盗阿福

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110;

int n, k;
int m[N], p[N], f[N];                   //f[i]表示在第i个点开餐馆的最大利润

int main()
{
    int T;
    cin >> T;

    while (T -- )
    {
        cin >> n >> k;
        for (int i = 1; i <= n; i ++ ) cin >> m[i];
        for (int i = 1; i <= n; i ++ ) cin >> p[i];

        for (int i = 1, j = 1, maxf = 0; i <= n; i ++ )
        {
            while (m[i] - m[j] > k)     //判断两家餐馆距离是否大于k
            {
                maxf = max(maxf, f[j]); //maxf储存i的上上个点,即(i - 2)点的最大值
                j ++ ;
            }

            f[i] = maxf + p[i];         //选了i点，(i - 1)点就不能选
        }

        int res = 0;                    //遍历所有状态找最大值
        for (int i = 1; i <= n; i ++ ) res = max(res, f[i]);

        cout << res << endl;
    }

    return 0;
}