description: Mr. Chanek has a new game called Dropping Balls. Initially, Mr. Chanek has a grid a of size n × m
  
  Each cell (x,y) contains an integer a_{x,y} denoting the direction of how the ball will move.
  
    * a_{x,y}=1 — the ball will move to the right (the next cell is (x, y + 1)); 
    * a_{x,y}=2 — the ball will move to the bottom (the next cell is (x + 1, y)); 
    * a_{x,y}=3 — the ball will move to the left (the next cell is (x, y - 1)). 
  
  
  
  Every time a ball leaves a cell (x,y), the integer a_{x,y} will change to 2. Mr. Chanek will drop k balls sequentially, each starting from the first row, and on the c_1, c_2, ..., c_k-th (1 ≤ c_i ≤ m) columns.
  
  Determine in which column each ball will end up in (position of the ball after leaving the grid).
  
  Input
  
  The first line contains three integers n, m, and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 10^5) — the size of the grid and the number of balls dropped by Mr. Chanek.
  
  The i-th of the next n lines contains m integers a_{i,1},a_{i,2},…,a_{i,m} (1 ≤ a_{i,j} ≤ 3). It will satisfy a_{i, 1} ≠ 3 and a_{i, m} ≠ 1.
  
  The next line contains k integers c_1, c_2, …, c_k (1 ≤ c_i ≤ m) — the balls' column positions dropped by Mr. Chanek sequentially.
  
  Output
  
  Output k integers — the i-th integer denoting the column where the i-th ball will end.
  
  Examples
  
  Input
  
  
  5 5 3
  1 2 3 3 3
  2 2 2 2 2
  2 2 2 2 2
  2 2 2 2 2
  2 2 2 2 2
  1 2 1
  
  
  Output
  
  
  2 2 1 
  
  
  Input
  
  
  1 2 2
  1 3
  1 2
  
  
  Output
  
  
  1 2 
  
  Note
  
  In the first example, the first ball will drop as follows. Note that the cell (1, 1) will change direction to the bottom direction.
  
  <image>
  
  The second and third balls will drop as follows. 
  
  <image>
  
  All balls will be dropped from the first row and on the c_1, c_2, ..., c_k-th columns respectively. A ball will stop dropping once it leaves the grid.
solution: #include <bits/stdc++.h>
  using namespace std;
  int arr[1005][1005];
  pair<int, int> dp[1005][1005];
  pair<int, int> findset(pair<int, int> x) {
    return (x == dp[x.first][x.second])
               ? x
               : dp[x.first][x.second] = findset(dp[x.first][x.second]);
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, m, h, i, j, k, a, b, q, d, c, cnt, cur2, sum = 0, sum2 = 0, t, maxn,
                                                    x = 0, y, z, ans, ans2, idx;
    string str = "", str2, str3;
    char ch;
    cin >> n >> m >> q;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        cin >> arr[i][j];
      }
    }
    for (i = 0; i < m; i++) dp[n][i] = make_pair(n, i);
    for (i = n - 1; i >= 0; i--) {
      for (j = 0; j < m; j++) {
        if (arr[i][j] == 2) {
          dp[i][j] = dp[i + 1][j];
        } else
          dp[i][j] = make_pair(i, j);
      }
    }
    for (i = 0; i < q; i++) {
      cin >> a;
      a--;
      x = 0;
      y = a;
      while (x != n) {
        if (arr[x][y] == 2) {
          pair<int, int> temp = findset(make_pair(x, y));
          x = temp.first;
          y = temp.second;
        } else if (arr[x][y] == 1) {
          arr[x][y] = 2;
          dp[x][y] = dp[x + 1][y];
          if (x && arr[x - 1][y] == 2) dp[x - 1][y] = dp[x + 1][y];
          y++;
        } else {
          arr[x][y] = 2;
          dp[x][y] = dp[x + 1][y];
          if (x && arr[x - 1][y] == 2) dp[x - 1][y] = dp[x + 1][y];
          y--;
        }
      }
      cout << y + 1 << " ";
    }
  }
  