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;
  ifstream in("in.in");
  ofstream out("out.out");
  const int nmax = 1005;
  int n, m, q, sol;
  int dp[2][nmax][nmax];
  vector<vector<int>> a;
  void calc(int i, int j) {
    dp[0][i][j] = dp[1][i][j] = 0;
    if (a[i][j] == 0) return;
    if (j < m - 1 && a[i][j + 1]) {
      dp[0][i][j] += dp[1][i][j + 1] + 1;
    }
    if (i < n - 1 && a[i + 1][j]) {
      dp[1][i][j] += dp[0][i + 1][j] + 1;
    }
  }
  void traverse(int i, int j, int dir) {
    if (i < 0 || j < 0) return;
    int oldDp = dp[0][i][j] + dp[1][i][j];
    calc(i, j);
    int newDp = dp[0][i][j] + dp[1][i][j];
    sol += newDp - oldDp;
    if (dir == 0) {
      traverse(i, j - 1, 1);
    } else {
      traverse(i - 1, j, 0);
    }
  }
  int main() {
    cin >> n >> m >> q;
    a = vector<vector<int>>(n, vector<int>(m, 1));
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        calc(i, j);
      }
    }
    sol = 0;
    int nrFree = n * m;
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        sol += dp[0][i][j] + dp[1][i][j];
      }
    }
    for (int i = 0; i < q; i++) {
      int y, x;
      cin >> y >> x;
      y--;
      x--;
      a[y][x] = 1 - a[y][x];
      if (a[y][x] == 0) {
        sol -= dp[0][y][x] + dp[1][y][x];
        dp[0][y][x] = dp[1][y][x] = 0;
        nrFree--;
      } else {
        calc(y, x);
        sol += dp[0][y][x] + dp[1][y][x];
        nrFree++;
      }
      traverse(y - 1, x, 0);
      traverse(y, x - 1, 1);
      cout << sol + nrFree << "\n";
    }
    return 0;
  }
  