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;
  struct yqh {
    int x, y;
  } fa[1020][1020];
  int n, m, k;
  int mp[1020][1020];
  yqh getfa(int x, int y) {
    if (fa[x][y].x == x && fa[x][y].y == y) return fa[x][y];
    fa[x][y] = getfa(fa[x][y].x, fa[x][y].y);
    return fa[x][y];
  }
  int get() {
    char t = getchar();
    while (t < '0' || t > '9') t = getchar();
    int x = 0;
    while (t >= '0' && t <= '9') {
      x *= 10;
      x += t - '0';
      t = getchar();
    }
    return x;
  }
  int main() {
    n = get();
    m = get();
    k = get();
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        mp[i][j] = get();
        if (mp[i][j] == 2) {
          fa[i][j].x = i + 1;
          fa[i][j].y = j;
        } else {
          fa[i][j].x = i;
          fa[i][j].y = j;
        }
      }
    }
    for (int j = 1; j <= m; ++j) {
      fa[n + 1][j].x = n + 1;
      fa[n + 1][j].y = j;
    }
    for (int i = 1; i <= k; ++i) {
      int sx = 1;
      int sy = get();
      while (sx <= n) {
        yqh nex = getfa(sx, sy);
        sx = nex.x;
        sy = nex.y;
        if (sx <= n) {
          int flag = mp[sx][sy];
          mp[sx][sy] = 2;
          fa[sx][sy].x = sx + 1;
          fa[sx][sy].y = sy;
          if (flag == 1)
            ++sy;
          else if (flag == 2)
            ++sx;
          else
            --sy;
        }
      }
      printf("%d", sy);
      if (i < k)
        printf(" ");
      else
        printf("\n");
    }
    return 0;
  }
  