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;
  const int N = 1000 + 5;
  int n, m, q, hor[N][N], ver[N][N], ans, x, y, ix, jx, v, h, vv, hh;
  int main() {
    vector<vector<bool> > free(N, vector<bool>(N, 1));
    scanf("%d%d%d", &n, &m, &q);
    ans = n * m;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        if (j > 0) {
          hor[i][j] = 1;
          if (i > 0) hor[i][j] += hor[i - 1][j - 1];
        }
      }
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        if (i > 0) {
          ver[i][j] = 1;
          if (j > 0) ver[i][j] += ver[i - 1][j - 1];
        }
      }
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        ans += hor[i][j];
        ans += ver[i][j];
        if (i > 0) ans += hor[i - 1][j];
        if (j > 0) ans += ver[i][j - 1];
      }
    while (q--) {
      scanf("%d%d", &x, &y);
      x--;
      y--;
      if (free[x][y]) {
        ans = ans - hor[x][y] - ver[x][y] - (x > 0 ? hor[x - 1][y] : 0) -
              (y > 0 ? ver[x][y - 1] : 0) - 1;
        h = hor[x][y], v = ver[x][y];
        free[x][y] = hor[x][y] = ver[x][y] = 0;
        ix = 0, jx = 1;
        while (free[x + ix][y + jx] && x + ix < n && y + jx < m) {
          if (jx > ix) {
            ans -= hor[x + ix][y + jx];
            ans -= v;
            hh = hor[x + ix][y + jx];
            if (jx == 1) hor[x + ix][y + jx] = 0;
            if (jx > 1) {
              hor[x + ix][y + jx] = 1;
              if (ix > 0) hor[x + ix][y + jx] += hor[x + ix - 1][y + jx - 1];
              ans += hor[x + ix][y + jx];
              ans += ver[x + ix][y + jx - 1];
            }
            ix++;
          } else {
            ans -= ver[x + ix][y + jx];
            ans -= hh;
            v = ver[x + ix][y + jx];
            if (ix > 0) {
              ver[x + ix][y + jx] = 1;
              if (jx > 0) ver[x + ix][y + jx] += ver[x + ix - 1][y + jx - 1];
              ans += ver[x + ix][y + jx];
              ans += hor[x + ix - 1][y + jx];
            }
            jx++;
          }
        }
        ix = 1, jx = 0;
        while (free[x + ix][y + jx] && x + ix < n && y + jx < m) {
          if (ix > jx) {
            ans -= ver[x + ix][y + jx];
            ans -= h;
            vv = ver[x + ix][y + jx];
            if (ix == 1) ver[x + ix][y + jx] = 0;
            if (ix > 1) {
              ver[x + ix][y + jx] = 1;
              if (jx > 0) ver[x + ix][y + jx] += ver[x + ix - 1][y + jx - 1];
              ans += ver[x + ix][y + jx];
              ans += hor[x + ix - 1][y + jx];
            }
            jx++;
          } else {
            ans -= hor[x + ix][y + jx];
            ans -= vv;
            h = hor[x + ix][y + jx];
            if (jx > 0) {
              hor[x + ix][y + jx] = 1;
              if (ix > 0) hor[x + ix][y + jx] += hor[x + ix - 1][y + jx - 1];
              ans += hor[x + ix][y + jx];
              ans += ver[x + ix][y + jx - 1];
            }
            ix++;
          }
        }
      } else {
        if (y > 0 && free[x][y - 1]) {
          hor[x][y] = 1;
          if (x > 0) hor[x][y] += hor[x - 1][y - 1];
        }
        if (x > 0 && free[x - 1][y]) {
          ver[x][y] = 1;
          if (y > 0) ver[x][y] += ver[x - 1][y - 1];
        }
        ans = ans + hor[x][y] + ver[x][y] + (x > 0 ? hor[x - 1][y] : 0) +
              (y > 0 ? ver[x][y - 1] : 0) + 1;
        v = h = 0;
        free[x][y] = 1;
        ix = 0, jx = 1;
        while (free[x + ix][y + jx] && x + ix < n && y + jx < m) {
          if (jx > ix) {
            ans -= hor[x + ix][y + jx];
            ans -= v;
            hh = hor[x + ix][y + jx];
            if (y + jx > 0) {
              hor[x + ix][y + jx] = 1;
              if (x + ix > 0) hor[x + ix][y + jx] += hor[x + ix - 1][y + jx - 1];
              ans += hor[x + ix][y + jx];
              ans += ver[x + ix][y + jx - 1];
            }
            ix++;
          } else {
            ans -= ver[x + ix][y + jx];
            ans -= hh;
            v = ver[x + ix][y + jx];
            if (x + ix > 0) {
              ver[x + ix][y + jx] = 1;
              if (y + jx > 0) ver[x + ix][y + jx] += ver[x + ix - 1][y + jx - 1];
              ans += ver[x + ix][y + jx];
              ans += hor[x + ix - 1][y + jx];
            }
            jx++;
          }
        }
        ix = 1, jx = 0;
        while (free[x + ix][y + jx] && x + ix < n && y + jx < m) {
          if (ix > jx) {
            ans -= ver[x + ix][y + jx];
            ans -= h;
            vv = ver[x + ix][y + jx];
            if (x + ix > 0) {
              ver[x + ix][y + jx] = 1;
              if (y + jx > 0) ver[x + ix][y + jx] += ver[x + ix - 1][y + jx - 1];
              ans += ver[x + ix][y + jx];
              ans += hor[x + ix - 1][y + jx];
            }
            jx++;
          } else {
            ans -= hor[x + ix][y + jx];
            ans -= vv;
            h = hor[x + ix][y + jx];
            if (y + jx > 0) {
              hor[x + ix][y + jx] = 1;
              if (x + ix > 0) hor[x + ix][y + jx] += hor[x + ix - 1][y + jx - 1];
              ans += hor[x + ix][y + jx];
              ans += ver[x + ix][y + jx - 1];
            }
            ix++;
          }
        }
      }
      printf("%d\n", ans);
    }
    return 0;
  }
  