算法1 BFS

Time Complexity: $O(N * Q)$ 至多10的6次方

#include <iostream>
#include <vector>
#include <queue>

using namespace std;
int n, l, k, x;

int bfs(vector<vector<int>>& g, int s) {

  queue<int> q{{s}};
  vector<int> seen(n + 1);
  seen[s] = 1;

  int steps = 0, num = 0;
  while (not q.empty() && steps <= l) {
    size_t sz = q.size();
    while (sz--) {
      const int cur = q.front(); q.pop();
      ++num;
      for (const auto& nxt : g[cur]) {
        if (seen[nxt]) continue;
        q.emplace(nxt);
        seen[nxt] = 1;
      }
    }
    ++steps;
  }

  return num - 1;
}

// debug helper function
void printGraph(vector<vector<int>>& g) {
  for (int u = 1; u <= n; ++u) {
    printf("%d: [", u);
    for (const auto& v : g[u]) printf(" %d", v);
    printf("]\n");
  }
  putchar('\n');
}

int main(int argc, char** argv) {
  cin >> n >> l;
  // build the directed graph
  vector<vector<int>> g(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> k;
    while (k--) {
      cin >> x;
      g[x].emplace_back(i);
    }
  }

  cin >> k;
  while (k--) {
    cin >> x;
    cout << bfs(g, x) << endl;
  }

  return 0;
}

TODO 算法2：Floyd