제출 #219290

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
219290		rama_pang	Lamps (JOI19_lamps)	C++14	0 / 100	1100 ms	67012 KiB

lamp

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

const int INF = 1e7;

int N;
string A, B;

void Read() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);

  cin >> N >> A >> B;
}

int NaiveDP() {
  vector<array<array<int, 4>, 4>> dp(N + 1); // dp[pos][second_last_operation][last_operation]
  // dp[pos] means A[pos...N] = B[pos...N]
  // operation 0 = skip operation, costs 0
  // operation 1 = off operation, costs 1 if not (second_last == 1 and last == 2), otherwise costs 0 (but then new_second_last is considered operation 0, since we cannot cascade alternating 1 and 2)
  // operation 2 = on operation, costs 1 if not (second_last == 2 and last == 1), otherwise costs 0 (but then new_second_last is considered operation 0, since we cannot cascade alternating 1 and 2)
  // operation 3 = xor operation, costs 1 if not (second_last == 1 and last == 2 or vice versa), otherwise costs 0
  

  for (int pos = N - 1; pos >= 0; pos--) {
    for (int second_last = 0; second_last < 4; second_last++) {
      for (int last = 0; last < 4; last++) {
        int &res = dp[pos][second_last][last];
        int is12 = second_last == 1 && last == 2;
        int is21 = second_last == 2 && last == 1;
        
        // operation skip (0)
        if (A[pos] == B[pos]) {
          res = dp[pos + 1][last][0];
        } else {
          res = INF;
        }

        // operation off (1)
        for (int nxt = pos; nxt < N; nxt++) {
          if (B[nxt] != '0') break;
          res = min(res, 1 + dp[nxt + 1][last][1]);
          res = min(res, 1 - is12 + dp[nxt + 1][0][1]);
        }

        // operation on (2)
        for (int nxt = pos; nxt < N; nxt++) {
          if (B[nxt] != '1') break;
          res = min(res, 1 + dp[nxt + 1][last][2]);
          res = min(res, 1 - is21 + dp[nxt + 1][0][2]);
        }

        // operation xor (3)
        for (int nxt = pos; nxt < N; nxt++) {
          if (A[nxt] == B[nxt]) break;
          res = min(res, 1 - (is12 || is21) + dp[nxt + 1][last][3]);
        }
      }
    }
  }

  return dp[0][0][0];
}

int main() {
  Read();
  cout << NaiveDP() << "\n";
  return 0;
}

• Subtask 1: 0 / 6
• Subtask 2: 0 / 41
• Subtask 3: 0 / 4
• Subtask 4: 0 / 49