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 = 1005;
  long long down[N][N], rit[N][N], ans;
  int flag[N][N], n, m;
  void fun(int x, int y, int f) {
    int a = x, b = y;
    int xia, zuo;
    if (f == -1) {
      xia = rit[a][b] + 1;
      zuo = down[a][b] + 1;
      xia = xia * f;
      zuo = zuo * f;
      ans--;
      ans = ans - rit[a][b] - down[a][b];
      rit[a][b] = down[a][b] = 0;
    } else if (f == 1) {
      ans++;
      rit[a][b] = down[a][b + 1] + (flag[a][b + 1] != 1 && (b + 1 <= m));
      down[a][b] = rit[a + 1][b] + (flag[a + 1][b] != 1 && (a + 1 <= n));
      ans = ans + rit[a][b] + down[a][b];
      xia = rit[a][b] + 1;
      zuo = down[a][b] + 1;
    }
    int now = 0;
    while (1) {
      if (now == 0) {
        a = a - 1;
        if (a < 1 || flag[a][b] == 1) break;
        down[a][b] = down[a][b] + xia;
        ans = ans + xia;
      } else {
        b = b - 1;
        if (b < 1 || flag[a][b] == 1) break;
        rit[a][b] = rit[a][b] + xia;
        ans = ans + xia;
      }
      now = now ^ 1;
    }
    now = 1;
    a = x;
    b = y;
    while (1) {
      if (now == 0) {
        a = a - 1;
        if (a < 1 || flag[a][b] == 1) break;
        down[a][b] = down[a][b] + zuo;
        ans = ans + zuo;
      } else {
        b = b - 1;
        if (b < 1 || flag[a][b] == 1) break;
        rit[a][b] = rit[a][b] + zuo;
        ans = ans + zuo;
      }
      now = now ^ 1;
    }
    printf("%lld\n", ans);
  }
  void solve() {
    int i, j, q, x, y;
    scanf("%d%d%d", &n, &m, &q);
    for (i = n; i >= 1; --i) {
      for (j = m; j >= 1; --j) {
        if (i == n)
          down[i][j] == 0;
        else
          down[i][j] = down[i][j] + 1 + rit[i + 1][j];
        if (j == m)
          rit[i][j] == 0;
        else
          rit[i][j] = rit[i][j] + 1 + down[i][j + 1];
      }
    }
    ans = 0;
    for (i = 1; i <= n; ++i) {
      for (j = 1; j <= m; ++j) {
        ans = ans + down[i][j] + rit[i][j];
      }
    }
    ans = ans + n * m;
    for (i = 1; i <= q; ++i) {
      scanf("%d%d", &x, &y);
      if (flag[x][y] == 1) {
        flag[x][y] = 0;
        fun(x, y, 1);
      } else {
        flag[x][y] = 1;
        fun(x, y, -1);
      }
    }
    return;
  }
  int main() { solve(); }
  