2023 rcpc

2023/rcpc/d2/editorial/b-the-floor-is-lava-46.cpp

@@ -0,0 +1,161 @@
+#include <bits/stdc++.h>
+#warning That's the baby, that's not my baby
+ 
+typedef long long ll;
+ 
+using namespace std;
+ 
+/**
+ 
+Solutie:
+ 
+Rotim la 45, acum ne fixam unul dintre cele K puncte.
+Observam ca va face cel mult n + m - 1 (< 2 * max(n, m)) pasi =>
+ne vom uita pentru r pasi, care este intaltimea maxima la care putem ajunge.
+Practic avem un query de maxim pe un patrat din matrice => putem face rmq
+ 
+Complexitate temporala: O(N^2 * logN + K * N)
+ 
+**/
+ 
+const int NMAX = 750;
+int a[NMAX + 1][NMAX + 1];
+int b[2 * NMAX + 2][2 * NMAX + 2];
+int rmqLine[12][2 * NMAX + 1][2 * NMAX + 1];
+int rmqCol[12][2 * NMAX + 1][2 * NMAX + 1];
+int lg2[2 * NMAX + 1];
+ 
+int N;
+const int p2[14] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192};
+ 
+int queryLine (int x, int l, int r) {
+  l = max(l, 1);
+  r = min(r, N);
+  int k = lg2[r - l + 1];
+  return max(rmqLine[k][x][l], rmqLine[k][x][r - p2[k] + 1]);
+}
+ 
+int queryCol (int x, int l, int r) {
+  l = max(l, 1);
+  r = min(r, N);
+  int k = lg2[r - l + 1];
+  return max(rmqCol[k][x][l], rmqCol[k][x][r - p2[k] + 1]);
+ 
+}
+ 
+int answer[NMAX * NMAX + 1];
+int mini[2 * NMAX + 1];
+ 
+int main() {
+  ios_base::sync_with_stdio(false);
+  cin.tie(0);
+ 
+  int n, m, k;
+  cin >> n >> m >> k;
+ 
+  N = max(n, m);
+  int maxel = 0;
+ 
+  for (int i = 1; i <= n; i++) {
+    for (int j = 1; j <= m; j++) {
+      cin >> a[i][j];
+      b[i + j - 1][i - j + N] = a[i][j];
+      maxel = std::max(maxel, a[i][j]);
+    }
+  }
+ 
+  N <<= 1;
+  for (int i = 2; i <= N; i++) {
+    lg2[i] = lg2[i >> 1] + 1;
+  }
+ 
+ 
+  for (int i = 1; i <= N; i++) {
+    for (int j = 1; j <= N; j++) {
+      if (!b[i][j]) {
+        rmqLine[0][i][j] = rmqCol[0][j][i] = 1e9;
+      }
+      rmqLine[0][i][j] = rmqCol[0][j][i] = b[i][j];
+//            cout << setw(3) << b[i][j];
+    }
+//        cout << '\n';
+  }
+ 
+  for (int k = 1; p2[k] <= N; k++) {
+    for (int i = 1; i <= N; i++) {
+      for (int j = 1; j + p2[k] - 1 <= N; j++) {
+        rmqLine[k][i][j] = max(rmqLine[k - 1][i][j], rmqLine[k - 1][i][j + p2[k - 1]]);
+      }
+    }
+    for (int j = 1; j <= N; j++) {
+      for (int i = 1; i + p2[k] - 1 <= N; i++) {
+        rmqCol[k][j][i] = max(rmqCol[k - 1][j][i], rmqCol[k - 1][j][i + p2[k - 1]]);
+      }
+    }
+  }
+ 
+  for (int r = 1; r <= N; r++) {
+    mini[r] = 1e9;
+  }
+ 
+  while (k--) {
+    int x, y;
+    cin >> x >> y;
+    int i = x + y - 1;
+    int j = x - y + N / 2;
+//        cout << " ? " << b[i][j] << " - " << a[x][y] << '\n';
+    int cur = 0;
+    for (int r = 1; r <= N; r++) {
+      int x1 = i - r + 1, y1 = j - r + 1, x2 = i + r - 1, y2 = j + r - 1;
+      if (x1 > 0) {
+        cur = max(cur, queryLine(x1, y1, y2));
+      }
+      if (x2 <= N) {
+        cur = max(cur, queryLine(x2, y1, y2));
+      }
+      if (y1 > 0) {
+        cur = max(cur, queryCol(y1, x1, x2));
+      }
+      if (y2 <= N) {
+        cur = max(cur, queryCol(y2, x1, x2));
+      }
+      mini[r] = min(mini[r], cur);
+    }
+  }
+ 
+  for (int r = 1; r <= N; r++) {
+//        cout << " > " << mini[r] << '\n';
+    if (mini[r] <= n * m) {
+      answer[mini[r] + 1] = max(answer[mini[r] + 1], r);
+    }
+  }
+  for (int i = 1; i <= n * m; i++) {
+    answer[i] = max(answer[i], answer[i - 1]);
+    if (i > maxel) {
+      std::cout << "-1 ";
+    } else {
+      cout << answer[i] << ' ';
+    }
+  }
+ 
+  return 0;
+}
+ 
+ 
+/**
+ 
+6 5 6
+30 14 11 22 16
+7  5  6  3  23
+20 1  5  2  17
+21 1  4  2  6
+17 7  3  24 3
+26 25 13 14 10
+2 2
+3 2
+4 4
+5 5
+2 4
+4 3
+ 
+**/