提出 #27114943

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ソースコード 拡げる

#include <bits/stdc++.h>
using namespace std;
// === Debug macro starts here ===
int recur_depth = 0;
#ifdef DEBUG
#define dbg(x)                                                              \
  {                                                                         \
    ++recur_depth;                                                          \
    auto x_ = x;                                                            \
    --recur_depth;                                                          \
    cerr << string(recur_depth, '	') << "\033[91m" << __func__ << ":" \
         << __LINE__ << "\t" << #x << " = " << x_ << "\033[39m" << endl;    \
  }
#else
#define dbg(x)
#endif
template <typename Ostream, typename Cont>
typename enable_if<is_same<Ostream, ostream>::value, Ostream&>::type operator<<(
    Ostream& os, const Cont& v) {
  os << "[";
  for (auto& x : v) {
    os << x << ", ";
  }
  return os << "]";
}
template <typename Ostream, typename... Ts>
Ostream& operator<<(Ostream& os, const pair<Ts...>& p) {
  return os << "{" << p.first << ", " << p.second << "}";
}
// === Debug macro ends here ===
#define REP(i, n) for (int i = 0; i < (int)(n); ++(i))
#define REP3(i, s, n) for (int i = (s); i < (int)(n); ++(i))
#define REP_R(i, n) for (int i = (int)(n)-1; (i) >= 0; --(i))
#define REP3R(i, m, n) for (int i = (int)(n)-1; (i) >= (int)(m); --(i))
#define ALL(x) std::begin(x), std::end(x)
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
#define MOD 1000000007
typedef long long ll;
typedef vector<int> vi;
ll dfs(ll n, vector<ll> T, vector<ll> K, vector<vector<ll>> A, vi done) {
  ll ans{};
  REP_R(i, K[n]) {
    if (done[A[n][i] - 1] == 0) ans += dfs(A[n][i] - 1, T, K, A, done);
  }
  done[n] = 1;
  ans += T[n];
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll N;
  cin >> N;
  vector<ll> T(N), K(N);
  vector<vector<ll>> A(N);
  REP(i, N) {
    cin >> T[i] >> K[i];
    vector<ll> b(K[i]);
    REP(j, K[i]) cin >> b[j];
    A[i] = b;
  }

  vi done(N, 0);
  ll ans = dfs(N - 1, T, K, A, done);
  cout << ans << '\n';

  return 0;
}

提出情報

ジャッジ結果