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>
  const int MAX_N = 1e3 + 10, dx[3] = {0, 1, 0}, dy[3] = {1, 0, -1};
  struct data {
    int x, y;
  } f[MAX_N][MAX_N];
  int n, m, q, mp[MAX_N][MAX_N];
  void merge(int x, int y) {
    mp[x][y] = 1;
    f[x][y] = (data){x + 1, y};
  }
  data findroot(int x, int y) {
    data nw = f[x][y];
    if (nw.x == -1)
      return (data){x, y};
    else
      return (f[x][y] = findroot(nw.x, nw.y));
  }
  int main() {
    scanf("%d %d %d", &n, &m, &q);
    for (int i = 1; i <= n + 1; ++i)
      for (int j = 1; j <= m; ++j) f[i][j] = (data){-1, -1};
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j) {
        scanf("%d", &mp[i][j]);
        --mp[i][j];
        if (mp[i][j] == 1) merge(i, j);
      }
    for (; q--;) {
      int x = 1, y;
      scanf("%d", &y);
      for (; x <= n;) {
        data nw = findroot(x, y);
        x = nw.x;
        y = nw.y;
        if (x <= n) {
          int tx = x + dx[mp[x][y]], ty = y + dy[mp[x][y]];
          merge(x, y);
          x = tx;
          y = ty;
        }
      }
      printf("%d ", y);
    }
    return 0;
  }
  