Submission #221516

#TimeUsernameProblemLanguageResultExecution timeMemory
221516oolimryRectangles (IOI19_rect)C++14
72 / 100
5031 ms834092 KiB
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#define all(x) x.begin(), x.end()

using namespace std;
typedef pair<int,int> ii;

int Left[2505][2505]; ///Left[r][c]: Left boundary from (r,c) provided (r,c) < (Left[r][c],c)
int Right[2505][2505]; ///Right[r][c]: Right boundary from (r,c) provided (r,c) < (Right[r][c],c)
static vector<int> Up[2505][2505]; ///Up[r][c]: All upper boundary from (r,c);

static vector<int> LeftRight[2505][2505]; ///LeftRight[L][R]: rows that [L,R] form a good pair
static vector<int> UpDown[2505][2505]; ///Updown[T][D]: columns that [T,D] form a good pair

inline int verify(int T, int B, int L, int R){ ///check if the rectangle enclosed by T, B, L, R is good (T, B, L, R are excluded)
  int ans = 1;

  ///number of rows between T and B (both exclusive) that form good pairs (L,R)
  if(LeftRight[L][R].empty() || lower_bound(all(LeftRight[L][R]), B) - upper_bound(all(LeftRight[L][R]), T) != (B - T - 1)) ans = 0;

  ///number of cols between L and R (both exclusive) that form good pairs (T,B)
  if(UpDown[T][B].empty() || lower_bound(all(UpDown[T][B]), R) - upper_bound(all(UpDown[T][B]), L) != (R - L - 1)) ans = 0;

  return ans;
}

long long count_rectangles(vector<vector<int> > arr) {
  int rows = arr.size();
  int cols = arr[0].size();

  for(int r = 0;r < rows;r++){
    for(int c = 0;c < cols;c++){
      Left[r][c] = -1;
      Right[r][c] = -1;
    }
  }

  for(int r = 0;r < rows;r++){
    vector<ii> s; s.push_back(ii(1e9,-1));
    //cout << r << ": \n";
    for(int i = 0;i < cols;i++){
      while(true){
        ii back = s.back();
        if(back.second != -1 && back.second != i - 1){
          int L = s.back().second;
          int R = i;
          LeftRight[L][R].push_back(r);

          if(arr[r][L] > arr[r][R]) Left[r][R] = L;
          else Right[r][L] = R;
        }
        if(s.back().first >= arr[r][i]){
          if(s.back().first == arr[r][i]) s.pop_back();
          break;
        }
        s.pop_back();
      }
      s.push_back(ii(arr[r][i],i));
    }
  }

  for(int c = 0;c < cols;c++){
    vector<ii> s; s.push_back(ii(1e9,-1));
    for(int i = 0;i < rows;i++){
      while(true){
        ii back = s.back();
        if(back.second != -1 && back.second != i - 1){
          int T = s.back().second;
          int B = i;

          UpDown[T][B].push_back(c);
          Up[B][c].push_back(T);
        }
        if(s.back().first >= arr[i][c]){
          if(s.back().first == arr[i][c]) s.pop_back();
          break;
        }
        s.pop_back();
      }
      s.push_back(ii(arr[i][c],i));
    }
  }

  int ans = 0;
  for(int r = 1;r < rows - 1;r++){
    for(int c = 1;c < cols - 1;c++){

      if(Right[r][c-1] != -1){
        int B = r + 1;
        int L = c - 1;
        int R = Right[r][L];
        for(int T : Up[B][c]){
          ans += verify(T,B,L,R);
        }
      }

      if(Left[r][c+1] != -1){
        int B = r + 1;
        int R = c + 1;
        int L = Left[r][R];
        for(int T : Up[B][c]){
          ans += verify(T,B,L,R);
        }
      }

    }
  }

  return ans;
}
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