Start by sorting the patches of the grass in increasing order of quality.

Let $f(i)$ be the maximum energy that we can accumulate if we end at patch $i$.

From patch $i$, we can compute the minimum distance from patch $i$ to every other patch. Then, for every patch $j$ where patch $j$ has lower quality grass than patch $i$, we have that $f(i) \ge f(j) + q_j - E * d(i,j)$.

It takes linear time to compute this information for a given patch. If we sort all the patches initially in $O(N \log N)$, then this process takes $O(N^2)$, which will run in time.

Here is Mark Gordon's code.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

#define MAXN 1010

int Q[MAXN];
int DP[MAXN];
int D[MAXN];
vector<int> E[MAXN];

int main() {
  freopen("buffet.in", "r", stdin);
  freopen("buffet.out", "w", stdout);

  int N, ECST;
  cin >> N >> ECST;
  for (int i = 0; i < N; i++) {
    int D;
    cin >> Q[i] >> D;
    for (int j = 0; j < D; j++) {
      int v;
      cin >> v;
      E[i].push_back(v - 1);
    }
  }

  vector<int> PI;
  for (int i = 0; i < N; i++) {
    PI.push_back(i);
  }
  sort(PI.begin(), PI.end(), [&](int x, int y) {
    return Q[x] < Q[y];
  });

  int result = 0;
  for (int i = N - 1; i >= 0; i--) {
    int u = PI[i];

    queue<int> q;
    memset(D, -1, sizeof(D));
    q.push(u);
    D[u] = 0;
    while (!q.empty()) {
      int v = q.front();
      q.pop();
      for (int i = 0; i < E[v].size(); i++) {
        int nv = E[v][i];
        if (D[nv] == -1) {
          D[nv] = D[v] + 1;
          q.push(nv);
        }
      }
    }

    int res = Q[u];
    for (int j = 0; j < N; j++) {
      if (D[j] != -1) {
        res = max(res, Q[u] + DP[j] - ECST * D[j]);
      }
    }
    DP[u] = res;
    result = max(result, res);
  }

  cout << result << endl;
  return 0;
}