description: You are given a matrix, consisting of n rows and m columns. The rows are numbered top to bottom, the columns are numbered left to right.
  
  Each cell of the matrix can be either free or locked.
  
  Let's call a path in the matrix a staircase if it: 
  
    * starts and ends in the free cell; 
    * visits only free cells; 
    * has one of the two following structures: 
      1. the second cell is 1 to the right from the first one, the third cell is 1 to the bottom from the second one, the fourth cell is 1 to the right from the third one, and so on; 
      2. the second cell is 1 to the bottom from the first one, the third cell is 1 to the right from the second one, the fourth cell is 1 to the bottom from the third one, and so on. 
  
  
  
  In particular, a path, consisting of a single cell, is considered to be a staircase.
  
  Here are some examples of staircases:
  
  <image>
  
  Initially all the cells of the matrix are free.
  
  You have to process q queries, each of them flips the state of a single cell. So, if a cell is currently free, it makes it locked, and if a cell is currently locked, it makes it free.
  
  Print the number of different staircases after each query. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.
  
  Input
  
  The first line contains three integers n, m and q (1 ≤ n, m ≤ 1000; 1 ≤ q ≤ 10^4) — the sizes of the matrix and the number of queries.
  
  Each of the next q lines contains two integers x and y (1 ≤ x ≤ n; 1 ≤ y ≤ m) — the description of each query.
  
  Output
  
  Print q integers — the i-th value should be equal to the number of different staircases after i queries. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.
  
  Examples
  
  Input
  
  
  2 2 8
  1 1
  1 1
  1 1
  2 2
  1 1
  1 2
  2 1
  1 1
  
  
  Output
  
  
  5
  10
  5
  2
  5
  3
  1
  0
  
  
  Input
  
  
  3 4 10
  1 4
  1 2
  2 3
  1 2
  2 3
  3 2
  1 3
  3 4
  1 3
  3 1
  
  
  Output
  
  
  49
  35
  24
  29
  49
  39
  31
  23
  29
  27
  
  
  Input
  
  
  1000 1000 2
  239 634
  239 634
  
  
  Output
  
  
  1332632508
  1333333000
solution: #include <bits/stdc++.h>
  using namespace std;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m, q;
    cin >> n >> m >> q;
    vector<vector<int>> a(n, vector<int>(m, 1));
    vector<vector<int>> b(n, vector<int>(m));
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        if (a[i][j]) {
          b[i][j] = max(b[i][j], i + j);
          if (i && j && a[i - 1][j - 1]) {
            b[i - 1][j - 1] = b[i][j];
          }
        }
      }
    }
    long long ans = n * m;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (!a[i][j]) {
          continue;
        }
        if (i) {
          if (a[i - 1][j]) {
            int A = (b[i][j] - i - j) / 2 + 1;
            int B = (b[i - 1][j] - i + 1 - j) / 2 + 1;
            if (A == B) {
              ans += A + B - 1;
            } else if (A < B) {
              ans += A * 2;
            } else {
              ans += B * 2 - 1;
            }
          }
        }
        if (j + 1 < m) {
          if (a[i][j + 1]) {
            int A = (b[i][j] - i - j) / 2 + 1;
            int B = (b[i][j + 1] - i - j - 1) / 2 + 1;
            if (A == B) {
              ans += A + B - 1;
            } else if (A > B) {
              ans += B * 2;
            } else {
              ans += A * 2 - 1;
            }
          }
        }
      }
    }
    for (int i = 0; i < q; i++) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      int oldx = x, oldy = y;
      int p = min(n - x - 1, m - y - 1);
      x += p;
      y += p;
      vector<pair<int, int>> tmp;
      while (x >= 0 && y >= 0) {
        if (x + 1 < n && y == m - 1) {
          tmp.push_back({x + 1, y});
        }
        if (y - 1 >= 0) {
          tmp.push_back({x, y - 1});
        }
        tmp.push_back({x, y});
        x--;
        y--;
      }
      for (auto j : tmp) {
        b[j.first][j.second] = 0;
      }
      for (auto j : tmp) {
        if (a[j.first][j.second]) {
          b[j.first][j.second] = max(b[j.first][j.second], j.first + j.second);
          if (j.first && j.second && a[j.first - 1][j.second - 1]) {
            b[j.first - 1][j.second - 1] = b[j.first][j.second];
          }
        }
      }
      for (auto j : tmp) {
        if (!a[j.first][j.second]) {
          continue;
        }
        if (j.first) {
          if (a[j.first - 1][j.second]) {
            int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
            int B = (b[j.first - 1][j.second] - j.first + 1 - j.second) / 2 + 1;
            if (A == B) {
              ans -= A + B - 1;
            } else if (A < B) {
              ans -= A * 2;
            } else {
              ans -= B * 2 - 1;
            }
          }
        }
        if (j.second + 1 < m) {
          if (a[j.first][j.second + 1]) {
            int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
            int B = (b[j.first][j.second + 1] - j.first - j.second - 1) / 2 + 1;
            if (A == B) {
              ans -= A + B - 1;
            } else if (A > B) {
              ans -= B * 2;
            } else {
              ans -= A * 2 - 1;
            }
          }
        }
      }
      if (a[oldx][oldy] == 1) {
        ans--;
      } else {
        ans++;
      }
      a[oldx][oldy] ^= 1;
      for (auto j : tmp) {
        b[j.first][j.second] = 0;
      }
      for (auto j : tmp) {
        if (a[j.first][j.second]) {
          b[j.first][j.second] = max(b[j.first][j.second], j.first + j.second);
          if (j.first && j.second && a[j.first - 1][j.second - 1]) {
            b[j.first - 1][j.second - 1] = b[j.first][j.second];
          }
        }
      }
      for (auto j : tmp) {
        if (!a[j.first][j.second]) {
          continue;
        }
        if (j.first) {
          if (a[j.first - 1][j.second]) {
            int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
            int B = (b[j.first - 1][j.second] - j.first + 1 - j.second) / 2 + 1;
            if (A == B) {
              ans += A + B - 1;
            } else if (A < B) {
              ans += A * 2;
            } else {
              ans += B * 2 - 1;
            }
          }
        }
        if (j.second + 1 < m) {
          if (a[j.first][j.second + 1]) {
            int A = (b[j.first][j.second] - j.first - j.second) / 2 + 1;
            int B = (b[j.first][j.second + 1] - j.first - j.second - 1) / 2 + 1;
            if (A == B) {
              ans += A + B - 1;
            } else if (A > B) {
              ans += B * 2;
            } else {
              ans += A * 2 - 1;
            }
          }
        }
      }
      cout << ans << endl;
    }
  }
  