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 maxn = 1e5 + 10;
  const int inf = 0x3f3f3f3f;
  const int mod = 1e9 + 7;
  const double PI = acos(-1.0);
  int dp[1111][1111][2];
  int mx[2] = {0, 1};
  int my[2] = {1, 0};
  struct node {
    int x, y, step, dir, type;
  };
  int n, m, q;
  int mp[1111][1111];
  int bfs(int x, int y, int type, int dir) {
    int step = 1;
    while (1) {
      if (type)
        y += dir;
      else
        x += dir;
      if (x >= 1 && x <= n && y >= 1 && y <= m && !mp[x][y])
        step++, type ^= 1;
      else
        return step;
    }
  }
  void solve() {
    cin >> n >> m >> q;
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        dp[i][j][0] = dp[i][j - 1][1] + 1;
        dp[i][j][1] = dp[i - 1][j][0] + 1;
        ans += dp[i][j][0] + dp[i][j][1] - 1;
      }
    }
    while (q--) {
      int x, y;
      scanf("%d %d", &x, &y);
      int cnt1 = bfs(x, y, 1, 1), cnt2 = bfs(x, y, 0, -1);
      int cnt3 = bfs(x, y, 0, 1), cnt4 = bfs(x, y, 1, -1);
      mp[x][y] ^= 1;
      int now = cnt1 * cnt2 + cnt3 * cnt4 - 1;
      if (mp[x][y])
        ans -= now;
      else
        ans += now;
      printf("%lld\n", ans);
    }
  }
  int main(int argc, char const *argv[]) {
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  