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 read() {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x / 10);
    putchar(x % 10 + '0');
  }
  template <typename T>
  void print(T x, char let) {
    print(x), putchar(let);
  }
  const int N = 1e3 + 5;
  int n, m, q, a[N][N], l[N][N][3];
  long long sum;
  void change(int x, int y, int c) {
    a[x][y] = c;
    if (c == 1) {
      sum--;
      for (int i = 0; i <= min(x - 1, y - 1); i++) {
        sum -= l[x - i][y - i][1] + l[x - i][y - i][2];
        l[x - i][y - i][1] = min(l[x - i][y - i][1], max(2 * i - 1, 0));
        l[x - i][y - i][2] = min(l[x - i][y - i][2], max(2 * i - 1, 0));
        sum += l[x - i][y - i][1] + l[x - i][y - i][2];
      }
      y--;
      for (int i = 0; i <= min(x - 1, y - 1); i++) {
        sum -= l[x - i][y - i][1];
        l[x - i][y - i][1] = min(l[x - i][y - i][1], 2 * i);
        sum += l[x - i][y - i][1];
      }
      y++;
      x--;
      for (int i = 0; i <= min(x - 1, y - 1); i++) {
        sum -= l[x - i][y - i][2];
        l[x - i][y - i][2] = min(l[x - i][y - i][2], 2 * i);
        sum += l[x - i][y - i][2];
      }
      cout << sum << endl;
    } else {
      sum++;
      for (int i = 0; i <= min(x - 1, y - 1); i++) {
        sum -= l[x - i][y - i][1] + l[x - i][y - i][2];
        if (a[x - i][y - i] == 0 && l[x - i][y - i][1] == max(2 * i - 1, 0)) {
          l[x - i][y - i][1] += l[x][y + 1][2] + 1 + (a[x][y + 1] == 1 ? 0 : 1);
        }
        if (a[x - i][y - i] == 0 && l[x - i][y - i][2] == max(2 * i - 1, 0)) {
          l[x - i][y - i][2] += l[x + 1][y][1] + 1 + (a[x + 1][y] == 1 ? 0 : 1);
        }
        sum += l[x - i][y - i][1] + l[x - i][y - i][2];
      }
      sum--;
      sum--;
      l[x][y][2]--;
      l[x][y][1]--;
      y--;
      for (int i = 0; i <= min(x - 1, y - 1); i++) {
        sum -= l[x - i][y - i][1];
        if (a[x - i][y - i] == 0 && l[x - i][y - i][1] == 2 * i) {
          l[x - i][y - i][1] +=
              l[x + 1][y + 1][1] + 1 + (a[x + 1][y + 1] == 1 ? 0 : 1);
        }
        sum += l[x - i][y - i][1];
      }
      y++;
      x--;
      for (int i = 0; i <= min(x - 1, y - 1); i++) {
        sum -= l[x - i][y - i][2];
        if (a[x - i][y - i] == 0 && l[x - i][y - i][2] == 2 * i) {
          l[x - i][y - i][2] +=
              l[x + 1][y + 1][2] + 1 + (a[x + 1][y + 1] == 1 ? 0 : 1);
        }
        sum += l[x - i][y - i][2];
      }
      cout << sum << endl;
    }
  }
  void solve() {
    n = read();
    m = read();
    q = read();
    for (int i = 0; i <= n + 1; i++)
      for (int j = 0; j <= m + 1; j++) a[i][j] = 1;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        a[i][j] = 0;
        l[i][j][1] = 2 * min(n - i, m - j);
        if (m - j > n - i) l[i][j][1]++;
        l[i][j][2] = 2 * min(n - i, m - j);
        if (n - i > m - j) l[i][j][2]++;
        sum += l[i][j][1] + l[i][j][2] + 1;
      }
    }
    int x, y;
    while (q--) {
      x = read();
      y = read();
      if (a[x][y] == 0)
        change(x, y, 1);
      else
        change(x, y, 0);
    }
  }
  int main() {
    solve();
    return 0;
  }
  