Submission #1072917

#	Time	Username	Problem	Language	Result	Execution time	Memory
1072917		kunzaZa183	Tortoise (CEOI21_tortoise)	C++17	0 / 100	88 ms	211752 KiB

tortoise

#include <bits/stdc++.h>
using namespace std;
  const int maxn = 300;
  int dp[maxn][maxn][maxn][2];
int main() {
  int n;
  cin>>n;
  vector<int> vi(n);
  for(auto &a:vi)
    cin>>a;

  for(int i = 0; i< maxn;i++)
    for(int j = 0; j < maxn; j++)
      for(int k = 0; k < maxn;k++)
        for(int l = 0; l < 2; l++)
          dp[i][j][k][l] = -1;

  auto repmax = [&](int &cur,int x) {
    cur = max(cur, x);
  };

  //tim, pos, last, has
  if(vi[0] >= 1) {
    dp[0][0][1][1] = 1;
  }else
    dp[0][0][0][0] = 0;

  for(int i = 0; i < 3*n ;i++)
    for(int j = 0; j < n; j++)
      for(int k = 0;k <= n; k++) {

        // cout<<i<<' '<<j<<' '<<k<<' '<<dp[i][j][k][0]<<' '<<dp[i][j][k][1]<<'\n';

        if(dp[i][j][k][0]!=-1 ){
          //send forward
          if(j + 1 < n) {
            repmax(dp[i + 1][j+1][k][0], dp[i][j][k][0]);

            if(vi[j+1] >= 1 && k <= j + 1 && i + 1 <= 2*(j+1)) {
              repmax(dp[i+1][j+1][j + 2][1], dp[i][j][k][0] + 1);
            }
          }
          //send back
          if(j - 1>= 0) {
            repmax(dp[i + 1][j-1][k][0], dp[i][j][k][0]);

            if(vi[j-1] >= 1 && k <= j - 1 && i + 1 <= 2*(j-1)) {
              repmax(dp[i+1][j - 1][j][1], dp[i][j][k][0] + 1);
            }
          }
        }

        if(dp[i][j][k][1] != -1) {
          if(j + 1 < n) {
            if(vi[j+1] == -1) {
              repmax(dp[i+1][j + 1][k][0], dp[i][j][k][1]);
            }else
            repmax(dp[i+ 1][j + 1][k][1], dp[i][j][k][1]);
          }

          if(j - 1 < n) {
            if(vi[j - 1] == -1) {
              repmax(dp[i + 1][j - 1][k][0], dp[i][j][k][1]);
            }else
            repmax(dp[i+1][j-1][k][1], dp[i + 1][j - 1][k][1]);
          }
        }
      }

  int maxval = 0;
  for(int i = 0; i <= 2*n; i++)
    for(int j = 0; j < n; j++)
      for(int k = 0; k < n; k++) 
        maxval = max({maxval, dp[i][j][k][0], dp[i][j][k][1]});

  cout<<count(vi.begin(), vi.end(), 1) - maxval<<'\n';
}

• Subtask 1: 0 / 8
• Subtask 2: 0 / 10
• Subtask 3: 0 / 30
• Subtask 4: 0 / 25
• Subtask 5: 0 / 27