nowcoder xinjun与阴阳师

/*
看似是完全背包的题目
可以看成是稍微加了种类限制的01背包
所以三重循环解决
第一重是种类
第二重是容量
第三重是每一种中选择一个模式不断更新dp一维数组
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 1000 + 5;

int n, m, dp[N];
vector <int> a[N], b[N];//a存贮价值，b存储花费

int main()
{
    int T;scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; ++i)
            a[i].clear(), b[i].clear();
        for (int i = 1; i <= n; ++i)
        {
            int x;scanf("%d", &x);
            for (int j = 1; j <= x; ++j)
            {
                int v;scanf("%d", &v);
                a[i].push_back(v);
            }
            for (int j = 1; j <= x; ++j)
            {
                int w;scanf("%d", &w);
                b[i].push_back(w);
            }
        }
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; ++i)
            for (int j = m; j >=0; j--)
                for (int k = 0; k < a[i].size(); ++k)
                    if (j >= b[i][k])
                        dp[j] = max(dp[j], dp[j - b[i][k]] + a[i][k]);

        printf("%d\n", dp[m]);
    }
    return 0;
}