不考虑具体怎么拿，只考虑每层拿到几个。
用f[i][j]存第i层拿了j个时的最大价值，求的方式是去掉一段连续的值之后剩余的最大价值。
要说的代码里都有注释了。

#include<bits/stdc++.h>
#define fi first
#define se second
#define pb(i) push_back(i)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=b;i>=a;i--)
#define mem(a,b) memset(a,b,sizeof(a))
#define VI vector<int>
#define VLL vector<ll>
#define MPII map<pair<int,int>,int>
#define VPII vector<pair<int,int>>
#define mp make_pair
#define PQI priority_queue<int>
#define lowbit(x) x&-x
#define debug(a) cout<<#a<<'='<<a<<'\n'
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e6+10;
const int INF = 0x3f3f3f3f;
const int inf = - INF;
const int mod = 1e9+7;
const double pi = acos(-1.0);
const double eps=1e-5;
vector<int> v[10005];
ll f[105][10005];//第i层拿j个的最大值
ll d[105][10005];//到第i层拿了j个
ll sum[105][100005];//每一层的前缀和，用于求f
int main(){
    int n, m, x, c;
    ll ans = 0;
    scanf("%d%d", &n, &m);
    for (int x, i = 1; i <= n; i++)
    {
        scanf("%d", &x);
        for (int t,j = 0; j < x; j++)
        {
            scanf("%d", &t);
            if(j > 0)sum[i][j] = sum[i][j - 1] + t;
            else sum[i][j] = t;
            v[i].push_back(t);
        }
    }
    for (int i = 1; i <= n; i++)
    {
        int sz = v[i].size();
        for (int j = 0; j <= min(m, sz); j++)
        {
            int t = v[i].size() - j;
            //k 从第k个开始挖
            //t 挖几个
            //printf("t = %d\n",t);
            ll& tmp = sum[i][v[i].size()-1];
            if(t > 0)
            {
                for (int k = 0; k + t - 1 < v[i].size();k++) {

                    if (k > 0)
                        f[i][j] = max(f[i][j], tmp - (sum[i][k + t - 1] - sum[i][k - 1]));
                    else
                        f[i][j] = max(f[i][j], tmp - sum[i][k + t - 1]);
                }
            }
            else
                f[i][j] = tmp;
            //printf("sum[%d] = %lld ,f[%d][%d] = %lld\n",i,sum[i][v[i].size()-1],i,j,f[i][j]);
        }
    }
    for (int i = 1; i <= n; i++)
    {
        //int sz = v[i].size();
        for (int j = 0; j <= m; j++) {
            for (int k = 0; k <= j; k++) {
                d[i][j] = max(d[i - 1][k] + f[i][j - k], d[i][j]);
                //printf("d[%d][%d] = %lld\n", i, j, d[i][j]);
            }
        }
    }
    printf("%lld\n",d[n][m]);
    return 0;
}