Submission #447401

#TimeUsernameProblemLanguageResultExecution timeMemory
447401dxz05Rectangles (IOI19_rect)C++14
50 / 100
1250 ms354708 KiB
#include "rect.h" #include <bits/stdc++.h> using namespace std; const int N = 2550; const int M = 222; bool good[M][M][M]; int n, m; vector<vector<int> > a; vector<int> dx = {0, 1, 0, -1}, dy = {1, 0, -1, 0}; bool visited[N][N]; int mnx, mny, mxx, mxy, cnt = 0; void dfs(int x, int y){ visited[x][y] = true; cnt++; mnx = min(mnx, x); mny = min(mny, y); mxx = max(mxx, x); mxy = max(mxy, y); for (int i = 0; i < 4; i++){ int X = x + dx[i], Y = y + dy[i]; if (X > 0 && X < n - 1 && Y > 0 && Y < m - 1 && !visited[X][Y] && a[X][Y] == 0) dfs(X, Y); } } long long count_rectangles(vector<vector<int> > _a) { a = _a; n = a.size(), m = a[0].size(); if (n < 3 || m < 3) return 0; if (n == 3) { int ans = 0; for (int l = 1; l < m - 1; l++) { int mx = 0; for (int r = l; r < m - 1; r++) { mx = max(mx, a[1][r]); if (a[1][r] >= min(a[0][r], a[2][r])) break; if (mx < min(a[1][l - 1], a[1][r + 1])) ans++; } } return ans; } if (n <= 200 && m <= 200) { for (int i = 1; i < n - 1; i++) { for (int x = 1; x < m - 1; x++) { int mx = 0; for (int y = x; y < m - 1; y++) { mx = max(mx, a[i][y]); if (mx < min(a[i][x - 1], a[i][y + 1])) { //cout << i << ' ' << x << ' ' << y << endl; good[i][x][y] = true; } } } } int ans = 0; for (int i = 1; i < n - 1; i++) { for (int x = 1; x < m - 1; x++) { for (int y = x; y < m - 1; y++) { int j = i; vector<int> v(m); while (j < n - 1 && good[j][x][y]) { int cnt = 0; for (int k = x; k <= y; k++) { v[k] = max(v[k], a[j][k]); cnt += v[k] < min(a[i - 1][k], a[j + 1][k]); } if (cnt == y - x + 1) ans++; j++; } } } } return ans; } int ans = 0; for (int i = 1; i < n - 1; i++){ for (int j = 1; j < m - 1; j++){ if (visited[i][j] || a[i][j] == 1) continue; mnx = mny = 1e9, mxx = mxy = cnt = 0; dfs(i, j); if (cnt != (mxx - mnx + 1) * (mxy - mny + 1)) continue; bool ok = true; for (int k = mnx; k <= mxx; k++){ if (a[k][mny - 1] == 0 || a[k][mxy + 1] == 0) ok = false; } for (int k = mny; k <= mxy; k++){ if (a[mnx - 1][k] == 0 || a[mxx + 1][k] == 0) ok = false; } if (ok) ans++; } } return ans; }
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