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;
  const int NMAX = 1000;
  int N, M, K;
  int a[NMAX + 2][NMAX + 2];
  struct state {
    int dad, sz;
    int down;
  };
  state ds[NMAX + 2][NMAX + 2];
  bool active[NMAX + 2][NMAX + 2];
  int root(int col, int p) {
    if (ds[p][col].dad != p) {
      return ds[p][col].dad = root(col, ds[p][col].dad);
    }
    return p;
  }
  void join(int col, int p, int q) {
    p = root(col, p);
    q = root(col, q);
    if (ds[p][col].sz > ds[q][col].sz) {
      swap(p, q);
    }
    ds[p][col].dad = q;
    ds[q][col].sz += ds[p][col].sz;
    ds[q][col].down = max(ds[q][col].down, ds[p][col].down);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> N >> M >> K;
    for (int i = 1; i <= N; i++) {
      for (int j = 1; j <= M; j++) {
        cin >> a[i][j];
        ds[i][j].dad = i;
        ds[i][j].sz = 1;
        ds[i][j].down = i;
      }
    }
    for (int i = 1; i <= N; i++) {
      for (int j = 1; j <= M; j++) {
        if (a[i][j] == 2) {
          active[i][j] = true;
          if (active[i - 1][j] == true) {
            join(j, i - 1, i);
          }
        }
      }
    }
    for (int i = 1; i <= K; i++) {
      int x = 1, y;
      cin >> y;
      while (true) {
        if (active[x][y] == true) {
          int r = root(y, x);
          x = ds[r][y].down + 1;
        }
        if (x >= N + 1) {
          cout << y << ' ';
          break;
        }
        int old_x = x, old_y = y;
        if (a[x][y] == 1) {
          y++;
        } else if (a[x][y] == 2) {
          x++;
        } else {
          y--;
        }
        a[old_x][old_y] = 2;
        active[old_x][old_y] = true;
        if (active[old_x - 1][old_y] == true) {
          join(old_y, old_x - 1, old_x);
        }
        if (active[old_x + 1][old_y] == true) {
          join(old_y, old_x + 1, old_x);
        }
      }
    }
    return 0;
  }
  