Submission #211675

#	Time	Username	Problem	Language	Result	Execution time	Memory
211675		EntityIT	Building 4 (JOI20_building4)	C++14	100 / 100	319 ms	26104 KiB

building4

#include<bits/stdc++.h>

using namespace std;

#define all(x) (x).begin(), (x).end()
#define sz(x) ( (int)(x).size() )
using LL = long long;

template<class T>
inline bool asMn(T &a, const T &b) { return a > b ? a = b, true : false; }
template<class T>
inline bool asMx(T &a, const T &b) { return a < b ? a = b, true : false; }

mt19937 rng( (uint32_t)chrono::steady_clock::now().time_since_epoch().count() );

pair<int, int> operator + (pair<int, int> &a, int b) { return { a.first + b, a.second + b }; }
void operator += (pair<int, int> &a, pair<int, int> b) {
  if (a == pair<int, int>{ -1, -1 } ) a = b;
  else {
    asMn(a.first, b.first);
    asMx(a.second, b.second);
  }
}

int main() {
  ios_base::sync_with_stdio(0); cin.tie(0);

  #ifdef FourLeafClover
  freopen("input", "r", stdin);
  #endif // FourLeafCLover

  int n; cin >> n;
  vector<int> a(n << 1), b(n << 1);
  for (int i = 0; i < n << 1; ++i) cin >> a[i];
  for (int i = 0; i < n << 1; ++i) cin >> b[i];

  vector<array<pair<int, int>, 2> > f(n << 1, { pair<int, int>{ -1, -1 }, pair<int, int>{ -1, -1 } });
  f[0] = { pair<int, int>{ 1, 1 }, pair<int, int>{ 0, 0 } };
  for (int i = 1; i < n << 1; ++i) {
    for (int j = 0; j < 2; ++j) if (~f[i - 1][j].first) {
      for (int k = 0; k < 2; ++k) if ( (j ? b[i - 1] : a[i - 1]) <= (k ? b[i] : a[i]) ) {
        f[i][k] += f[i - 1][j] + !k;
      }
    }
//    for (int j = 0; j < 2; ++j) cerr << f[i][j].first << '-' << f[i][j].second << ' ';
//    cerr << '\n';
  }

  int curJ = -1;
  for (int j = 0; j < 2; ++j) if (f.back()[j].first <= n && f.back()[j].second >= n) curJ = j;

  if (!~curJ) return cout << "-1\n", 0;

  string ans(n << 1, ' ');
  for (int i = (n << 1) - 1, cur = n; i; --i) {
//    cerr << "i: " << i << "  curJ: " << curJ << '\n';
    ans[i] = curJ ? 'B' : 'A';
    cur = cur + (!curJ ? -1 : 0);
    for (int nxtJ = 0; nxtJ < 2; ++nxtJ) if (~f[i - 1][nxtJ].first
                                             && (curJ ? b[i] : a[i]) >= (nxtJ ? b[i - 1] : a[i - 1])
                                             && f[i - 1][nxtJ].first <= cur && cur <= f[i - 1][nxtJ].second) {
      curJ = nxtJ;
      break ;
    }
  }
  ans[0] = curJ ? 'B' : 'A';

  cout << ans << '\n';

  return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 89 / 89