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() {
    int n, m, q;
    cin >> n >> m >> q;
    vector<int> x(q), y(q);
    for (int i = (0); i < (q); i++) {
      cin >> x[i] >> y[i];
      x[i]--;
      y[i]--;
    }
    vector<vector<long long>> u(n, vector<long long>(m, 1)),
        l(n, vector<long long>(m, 1));
    long long ans = 0, t = 0;
    for (int i = (0); i < (n); i++) {
      for (int j = (0); j < (m); j++) {
        if (i > 0) u[i][j] = l[i - 1][j] + 1;
        if (j > 0) l[i][j] = u[i][j - 1] + 1;
        ans += u[i][j] + l[i][j];
        t++;
      }
    }
    for (int i = (0); i < (q); i++) {
      if (u[x[i]][y[i]] > 0) {
        ans -= u[x[i]][y[i]] + l[x[i]][y[i]];
        t--;
        u[x[i]][y[i]] = 0;
        l[x[i]][y[i]] = 0;
      } else {
        t++;
        u[x[i]][y[i]] = 1;
        l[x[i]][y[i]] = 1;
        if (x[i] > 0) u[x[i]][y[i]] = l[x[i] - 1][y[i]] + 1;
        if (y[i] > 0) l[x[i]][y[i]] = u[x[i]][y[i] - 1] + 1;
        ans += u[x[i]][y[i]] + l[x[i]][y[i]];
      }
      for (int j = (x[i]); j < (n); j++) {
        for (int k = (y[i] + j - x[i] - 1); k < (y[i] + j - x[i] + 2); k++) {
          if (j == x[i]) {
            if (k < y[i] + j - x[i] + 1) continue;
          }
          if (k >= m) continue;
          if (u[j][k] == 0) continue;
          ans -= u[j][k] + l[j][k];
          u[j][k] = 1;
          l[j][k] = 1;
          if (j > 0) u[j][k] = l[j - 1][k] + 1;
          if (k > 0) l[j][k] = u[j][k - 1] + 1;
          ans += u[j][k] + l[j][k];
        }
      }
      cout << ans - t << endl;
    }
  }
  