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>
  template <class T>
  long long sqr(T a) {
    return 1LL * a * a;
  }
  template <class T>
  void compress(T &a) {
    sort(a.begin(), a.end());
    a.resize(unique(all(a)) - a.begin());
  }
  using namespace std;
  const int N = 1e3 + 10;
  const int M = log2(N) + 10;
  const int MOD = 1e9 + 7;
  const int INF = 1e9;
  const long long INFI = 1e18;
  const double long pi = acos(-1.00);
  const double long eps = 1e-9;
  int dist[N][N], dp[N][N], sta[N][N], n, m, q, res = 0;
  int calc(int x, int y, int z) {
    if (y > z) {
      swap(y, z);
    }
    if (x <= y) {
      if (!x) {
        return 0;
      }
      return (4 * x - 1);
    }
    if (z < x) {
      return (4 * y + 1 + 2 * (z - y));
    }
    return (4 * y + 1 + 2 * (x - y - 1) + 1);
  }
  void update_dp(int x, int y) {
    for (int i = 0; i < min(n, m); i++) {
      if (x - i > 0 && y - i > 0) {
        int cur = dp[x - i][y - i];
        dp[x - i][y - i] = calc(dist[x - i][y - i], dist[x - i + 1][y - i],
                                dist[x - i][y - i + 1]);
        res += dp[x - i][y - i] - cur;
      }
      if (x - i - 1 > 0 && y - i > 0) {
        int cur = dp[x - i - 1][y - i];
        dp[x - i - 1][y - i] = calc(dist[x - i - 1][y - i], dist[x - i][y - i],
                                    dist[x - i - 1][y - i + 1]);
        res += dp[x - i - 1][y - i] - cur;
      }
      if (x - i > 0 && y - i - 1 > 0) {
        int cur = dp[x - i][y - i - 1];
        dp[x - i][y - i - 1] =
            calc(dist[x - i][y - i - 1], dist[x - i + 1][y - i - 1],
                 dist[x - i][y - i]);
        res += dp[x - i][y - i - 1] - cur;
      }
    }
  }
  void update_0(int x, int y) {
    for (int i = 0; i < min(n, m); i++) {
      if (x == i || y == i) {
        break;
      }
      if (dist[x - i][y - i] == i) {
        dist[x - i][y - i] = dist[x + 1][y + 1] + i + 1;
      }
    }
    update_dp(x, y);
  }
  void update_1(int x, int y) {
    for (int i = 0; i < min(n, m); i++) {
      if (x == i || y == i) {
        break;
      }
      dist[x - i][y - i] = min(dist[x - i][y - i], i);
    }
    update_dp(x, y);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        dist[i][j] = min(n - i, m - j) + 1;
      }
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        dp[i][j] = calc(dist[i][j], dist[i + 1][j], dist[i][j + 1]);
        res += dp[i][j];
      }
    }
    while (q--) {
      int x, y;
      cin >> x >> y;
      sta[x][y] ^= 1;
      if (!sta[x][y]) {
        update_0(x, y);
      } else {
        update_1(x, y);
      }
      cout << res << "\n";
    }
    return 0;
  }
  