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;
  int t[2002][2002];
  void inc(int d, int i, int delta, int n) {
    for (; i <= n; i |= i + 1) t[d][i] += delta;
  }
  int sum(int d, int i) {
    int res = 0;
    for (; i > 0; i = (i & (i + 1)) - 1) res += t[d][i];
    return res;
  }
  bool isFree(int n, int l, int r) { return sum(n, r) == sum(n, l - 1); }
  bool a[1001][1001];
  int n, m;
  int diagonal(int x, int y) { return (n - x) + y; }
  int diagonal_index(int x, int y) {
    int k = diagonal(x, y);
    if (k <= n)
      return y;
    else
      return x;
  }
  int diagonal_size(int x, int y) {
    int k = diagonal(x, y);
    if (k <= n && k <= m) return k;
    if (k <= max(n, m)) return min(n, m);
    return min(n, m) - (k - max(n, m));
  }
  int get_past(int x, int y) {
    if (x > n || y > m || a[x][y]) return 0;
    int d = diagonal(x, y);
    int i = diagonal_index(x, y);
    int n = diagonal_size(x, y);
    int l = i, r = n, m;
    while (l != r) {
      m = (l + r + 1) / 2;
      if (isFree(d, i, m))
        l = m;
      else
        r = m - 1;
    }
    return l - i + 1;
  }
  int get_tepa(int x, int y) {
    if (x < 1 || y < 1 || a[x][y]) return 0;
    int d = diagonal(x, y);
    int i = diagonal_index(x, y);
    int l = 1, r = i, m;
    while (l != r) {
      m = (l + r) / 2;
      if (isFree(d, m, i))
        r = m;
      else
        l = m + 1;
    }
    return i - l + 1;
  }
  void t_main() {
    long long res = 0;
    int q, x, y;
    cin >> n >> m >> q;
    long long Q, W;
    for (int i = 1; i < n; i++) {
      for (int j = 1; j < m; j++) {
        res += (diagonal_size(i, j) - diagonal_index(i, j) + 1) +
               (diagonal_size(i + 1, j) - diagonal_index(i + 1, j) + 1);
        res += (diagonal_size(i, j) - diagonal_index(i, j) + 1) +
               (diagonal_size(i, j + 1) - diagonal_index(i, j + 1) + 1);
        res--;
      }
    }
    res += (n + m - 2) * 2 + 1;
    int dia, ind, sz, xyp, xyt, xpyp, xymt, xypp, xmyt;
    bool f;
    for (int i = 0; i < q; i++) {
      cin >> x >> y;
      dia = diagonal(x, y);
      ind = diagonal_index(x, y);
      sz = diagonal_size(x, y);
      if (a[x][y]) {
        f = true;
        inc(dia, ind, -1, sz);
        a[x][y] = 0;
      } else
        f = false;
      xyp = get_past(x, y);
      xyt = get_tepa(x, y);
      xpyp = min(get_past(x + 1, y), xyp);
      xymt = min(get_tepa(x, y - 1), xyt);
      xypp = min(get_past(x, y + 1), xyp);
      xmyt = min(get_tepa(x - 1, y), xyt);
      Q = min(xyp, xpyp + 1) + xpyp;
      Q += (min(xyt, xymt + 1) - 1 + xymt) * Q;
      W = min(xyp, xypp + 1) + xypp;
      W += (min(xyt, xmyt + 1) - 1 + xmyt) * W;
      Q = Q + W - 1;
      if (f) {
        res += Q;
      } else {
        a[x][y] = 1;
        inc(dia, ind, 1, sz);
        res -= Q;
      }
      cout << res << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) t_main();
    return 0;
  }
  