(Analysis by Nick Wu)

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;
}