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;}