Dynamic Programming! Suppose $$$T$$$ is the number of trees, $$$H$$$ is the height of each tree and $$$F$$$ is the parameter that is given in the input. At first let $$$tree[t][h]$$$ be the number of acorns on the tree $$$t$$$ at height $$$h$$$. Simply let $$$dp[t][h]$$$ be the maximum number of acorns the squirrel can collect if he is on the tree $$$t$$$ at height $$$h$$$. $$$dp[t][0] = 0$$$ for every $$$t$$$. Also let $$$best[h]$$$ be the maximum number of acorns he can collect, if he is at the height $$$h$$$ and he can freely choose any tree to descent from, in other words, $$$best[h]$$$ is the maximum of $$$dp[t][h]$$$ for all $$$t$$$. So the recurrence will be like this : $$$dp[t][h] = tree[t][h] + max(dp[t][h - 1], best[h - F])$$$. This means, the squirrel is either going to remain on the same tree and descent one meter, or is going to jump on another tree and lose $$$F$$$ meters in height, and we choose the maximum among those two cases. Note that $$$h - F$$$ could be negative so in that case it must be assumed zero. The answer to the problem is $$$best[H]$$$.

#include <bits/stdc++.h>
using namespace std;

void solve() {
  int T, H, F;
  cin >> T >> H >> F;
  vector<vector<int>> tree(T, vector<int>(H + 1));
  for(int t = 0; t < T; ++t) {
    int a; cin >> a;
    for(int j = 0; j < a; ++j) {
      int h; cin >> h;
      tree[t][h] += 1;
    }
  }
  vector<vector<int>> dp(T, vector<int>(H + 1));
  vector<int> best(H + 1);
  for(int h = 1; h <= H; ++h) {
    for(int t = 0; t < T; ++t) {
      dp[t][h] = tree[t][h] + max(dp[t][h - 1], (h - F >= 0 ? best[h - F] : 0));
      best[h] = max(best[h], dp[t][h]);
    }
  }
  cout << best[H] << '\n';
}

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int tc;
  cin >> tc;
  while(tc--) {
    solve();
  }
  cin >> tc;
}