제출 #679604

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
679604		peijar	Nautilus (BOI19_nautilus)	C++17	100 / 100	181 ms	416 KiB

nautilus

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

namespace std {
template <typename T> ostream &operator<<(ostream &out, const vector<T> &vec) {
  out << "[";
  for (int i = 0; i < (int)vec.size(); ++i) {
    out << vec[i];
    if (i + 1 < (int)vec.size())
      out << ", ";
  }
  return out << "]";
}
} // namespace std

void dbg_out() { cout << endl; }
template <typename Head, typename... Tail> void dbg_out(Head H, Tail... T) {
  cout << ' ' << H;
  dbg_out(T...);
}

#ifdef DEBUG
#define dbg(...) cout << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif

const int MAXN = 500;

signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);

  int nbLig, nbCol, nbOp;
  cin >> nbLig >> nbCol >> nbOp;

  vector<bitset<MAXN>> canUse(nbLig);
  for (int lig = 0; lig < nbLig; ++lig)
    for (int col = 0; col < nbCol; ++col) {
      char c;
      cin >> c;
      if (c == '.')
        canUse[lig][col] = 1;
    }

  vector<bitset<MAXN>> curPossible(nbLig);
  for (int lig = 0; lig < nbLig; ++lig)
    curPossible[lig] = canUse[lig];

  string s;
  cin >> s;
  for (char c : s) {
    vector<bitset<MAXN>> nxtPossible(nbLig);
    if (c == 'N' or c == '?') {
      for (int lig = 0; lig + 1 < nbLig; ++lig)
        nxtPossible[lig] |= curPossible[lig + 1];
    }
    if (c == 'S' or c == '?') {
      for (int lig = 1; lig < nbLig; ++lig)
        nxtPossible[lig] |= curPossible[lig - 1];
    }
    if (c == 'W' or c == '?') {
      for (int lig = 0; lig < nbLig; ++lig)
        nxtPossible[lig] |= curPossible[lig] >> 1;
    }
    if (c == 'E' or c == '?') {
      for (int lig = 0; lig < nbLig; ++lig)
        nxtPossible[lig] |= curPossible[lig] << 1;
    }
    for (int lig = 0; lig < nbLig; ++lig)
      curPossible[lig] = nxtPossible[lig] & canUse[lig];
  }
  int sol = 0;
  for (int lig = 0; lig < nbLig; ++lig)
    sol += curPossible[lig].count();
  cout << sol << endl;
}

• Subtask 1: 29 / 29
• Subtask 2: 37 / 37
• Subtask 3: 34 / 34