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 const N = 1e3 + 5;
  class dsu {
   public:
    int _n;
    std::vector<int> p;
    dsu() : _n(0){};
    explicit dsu(int n) : _n(n), p(n, -1) {}
    int merge(int a, int b) {
      assert(0 <= a && a < _n);
      assert(0 <= b && b < _n);
      int x = leader(a), y = leader(b);
      if (x == y) return x;
      p[x] += p[y];
      p[y] = x;
      return x;
    }
    void init(int n) {
      p.resize(n);
      for (int i = 1; i < n; i++) p[i] = i;
    }
    bool same(int a, int b) {
      assert(0 <= a && a < _n);
      assert(0 <= b && b < _n);
      return leader(a) == leader(b);
    }
    int leader(int a) {
      assert(0 <= a && a < _n);
      if (p[a] < 0) return a;
      return (p[a] == a) ? a : p[a] = leader(p[a]);
    }
    int size(int a) {
      assert(0 <= a && a < _n);
      return -p[leader(a)];
    }
    std::vector<std::vector<int>> groups() {
      std::vector<int> leader_buf(_n), group_size(_n);
      for (int i = 0; i < _n; i++) {
        leader_buf[i] = leader(i);
        group_size[leader_buf[i]]++;
      }
      std::vector<std::vector<int>> result(_n);
      for (int i = 0; i < _n; i++) result[i].reserve(group_size[i]);
      for (int i = 0; i < _n; i++) result[leader_buf[i]].push_back(i);
      result.erase(
          std::remove_if(result.begin(), result.end(),
                         [&](const std::vector<int> &v) { return v.empty(); }),
          result.end());
      return result;
    }
  };
  int g[N][N];
  int main() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<dsu> v(m + 2, dsu(n + 2));
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) cin >> g[i][j];
    }
    for (int i = 1; i <= m; i++) {
      v[i].init(n + 1);
      for (int j = 1; j <= n; j++) {
        if (g[j][i] == 2) v[i].p[j] = j + 1;
      }
    }
    while (k--) {
      int j;
      cin >> j;
      int i = v[j].leader(1);
      while (i <= n) {
        v[j].p[i] = i + 1;
        if (g[i][j] == 1)
          g[i][j++] = 2;
        else
          g[i][j--] = 2;
        i = v[j].leader(i);
        ;
      }
      cout << j << " ";
    }
  }
  