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;
  long long cnt[1005][1005][2];
  bool vis[1005][1005];
  long long n, m, ans;
  long long getadd(long long x, long long y, int dir) {
    if (x >= n || y >= m) return 0;
    if (vis[x][y]) return 0;
    if (dir == 0) return getadd(x, y + 1, 1) + 1;
    return getadd(x + 1, y, 0) + 1;
  }
  void dfs(long long x, long long y, long long k, int dir) {
    if (x < 0 || y < 0) return;
    if (vis[x][y]) return;
    if (dir == 0) {
      cnt[x][y][0] += k;
      ans += k;
      dfs(x - 1, y, k, 1);
    } else {
      cnt[x][y][1] += k;
      ans += k;
      dfs(x, y - 1, k, 0);
    }
  }
  int main() {
    long long i, j, q, x, y, k0, k1;
    scanf("%lld%lld%lld", &n, &m, &q);
    ans = 0;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        cnt[i][j][0] = min(n - i, m - j) * 2 - 1;
        cnt[i][j][1] = min(n - i, m - j) * 2 - 1;
        if (n - i > m - j) cnt[i][j][1]++;
        if (n - i < m - j) cnt[i][j][0]++;
        ans += cnt[i][j][0] + cnt[i][j][1] - 1;
      }
    }
    while (q--) {
      scanf("%lld%lld", &x, &y);
      if (vis[x - 1][y - 1]) {
        vis[x - 1][y - 1] = false;
        k1 = getadd(x - 1, y - 1, 0);
        k0 = getadd(x - 1, y - 1, 1);
        dfs(x - 1, y - 1, k1, 0);
        dfs(x - 1, y - 1, k0, 1);
        ans--;
      } else {
        k0 = cnt[x - 1][y - 1][0];
        k1 = cnt[x - 1][y - 1][1];
        cnt[x - 1][y - 1][0] = 0;
        cnt[x - 1][y - 1][1] = 0;
        ans -= k0 + k1 - 1;
        dfs(x - 1, y - 2, -k1, 0);
        dfs(x - 2, y - 1, -k0, 1);
        vis[x - 1][y - 1] = true;
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  