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 M = 1000000007;
  const int MM = 998244353;
  template <typename T, typename U>
  static inline void amin(T &x, U y) {
    if (y < x) x = y;
  }
  template <typename T, typename U>
  static inline void amax(T &x, U y) {
    if (x < y) x = y;
  }
  const int N = 1005;
  long long dp[N][N][4];
  int st[N][N];
  int _runtimeTerror_() {
    int n, m, Q;
    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][3] = dp[i][j - 1][2] + 1;
        dp[i][j][2] = dp[i - 1][j][3] + 1;
      }
    }
    for (int i = n; i >= 1; --i) {
      for (int j = m; j >= 1; --j) {
        dp[i][j][0] = dp[i][j + 1][1] + 1;
        dp[i][j][1] = dp[i + 1][j][0] + 1;
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        ans += dp[i][j][0] + dp[i][j][1] - 1;
      }
    }
    auto go = [&](int x, int y) {
      int it = 0;
      int i = x, j = y;
      auto upd1 = [&](int i, int j) {
        if (i < 1 || i > n || j < 1 || j > m) {
          return;
        }
        if (st[i][j] == 1) {
          return;
        }
        dp[i][j][3] = dp[i][j - 1][2] + 1;
        dp[i][j][2] = dp[i - 1][j][3] + 1;
      };
      while (true) {
        upd1(i + 1, j);
        upd1(i, j + 1);
        upd1(i + 1, j + 1);
        i += 1, j += 1;
        if (i > n || j > m) {
          break;
        }
      }
      auto upd2 = [&](int i, int j) {
        if (i < 1 || i > n || j < 1 || j > m) {
          return;
        }
        if (st[i][j] == 1) {
          return;
        }
        dp[i][j][0] = dp[i][j + 1][1] + 1;
        dp[i][j][1] = dp[i + 1][j][0] + 1;
      };
      i = x, j = y;
      while (true) {
        upd2(i - 1, j);
        upd2(i, j - 1);
        upd2(i - 1, j - 1);
        i -= 1, j -= 1;
        if (i < 1 || j < 1) {
          break;
        }
      }
    };
    auto update = [&](int x, int y) {
      if (st[x][y]) {
        int i = x, j = y;
        dp[i][j][3] = dp[i][j - 1][2] + 1;
        dp[i][j][2] = dp[i - 1][j][3] + 1;
        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][2];
        ans += dp[i][j][1] * dp[i][j][3];
        ans -= 1;
        go(x, y);
      } else {
        int i = x, j = y;
        ans -= dp[i][j][0] * dp[i][j][2];
        ans -= dp[i][j][1] * dp[i][j][3];
        ans += 1;
        dp[i][j][3] = 0;
        dp[i][j][2] = 0;
        dp[i][j][0] = 0;
        dp[i][j][1] = 0;
        go(x, y);
      }
      st[x][y] ^= 1;
    };
    for (int i = 1; i <= Q; ++i) {
      int x, y;
      cin >> x >> y;
      update(x, y);
      assert(ans >= 0);
      cout << ans << "\n";
    }
    return 0;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int TESTS = 1;
    while (TESTS--) _runtimeTerror_();
    return 0;
  }
  