description: Alice has an empty grid with n rows and m columns. Some of the cells are marked, and no marked cells are adjacent to the edge of the grid. (Two squares are adjacent if they share a side.) 
  
  Alice wants to fill each cell with a number such that the following statements are true: 
  
    * every unmarked cell contains either the number 1 or 4; 
    * every marked cell contains the sum of the numbers in all unmarked cells adjacent to it (if a marked cell is not adjacent to any unmarked cell, this sum is 0); 
    * every marked cell contains a multiple of 5. 
  
  Alice couldn't figure it out, so she asks Bob to help her. Help Bob find any such grid, or state that no such grid exists.
  
  Input
  
  The first line of input contains two integers n and m (1 ≤ n, m ≤ 500) — the number of rows and the number of columns in the grid, respectively.
  
  Then n lines follow, each containing m characters. Each of these characters is either '.' or 'X' — an unmarked and a marked cell, respectively. No marked cells are adjacent to the edge of the grid.
  
  Output
  
  Output "'NO" if no suitable grid exists. Otherwise, output "'YES"'. Then output n lines of m space-separated integers — the integers in the grid.
  
  Examples
  
  Input
  
  
  5 5
  .....
  .XXX.
  .X.X.
  .XXX.
  .....
  
  
  Output
  
  
  YES
  4 1 4 4 1
  4 5 5 5 1
  4 5 1 5 4
  1 5 5 5 4
  1 4 4 1 4
  
  
  Input
  
  
  5 5
  .....
  .XXX.
  .XXX.
  .XXX.
  .....
  
  
  Output
  
  
  NO
  
  
  Input
  
  
  3 2
  ..
  ..
  ..
  
  
  Output
  
  
  YES
  4 1
  4 1
  1 4
  
  
  Input
  
  
  9 9
  .........
  .XXXXX.X.
  .X...X...
  .X.XXXXX.
  .X.X.X.X.
  .X.XXX.X.
  .X.....X.
  .XXXXXXX.
  .........
  
  
  Output
  
  
  YES
  4 4 4 1 4 1 4 1 4
  1 5 5 5 5 5 4 10 1
  4 5 1 4 1 5 4 4 4
  4 5 1 5 5 0 5 5 1
  4 5 1 5 4 5 1 5 4
  4 5 1 5 5 5 4 5 1
  1 5 4 4 1 1 4 5 1
  4 5 5 5 5 5 5 5 4
  1 1 1 1 4 4 1 1 4
  
  Note
  
  It can be shown that no such grid exists for the second test.
solution: #include <bits/stdc++.h>
  using namespace std;
  vector<pair<int, int> > e[505][505];
  char c[505][505];
  int t[4][2] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}};
  int col[505][505];
  void dfs(int x, int y, int cl) {
    col[x][y] = cl;
    for (int i = 0; i < e[x][y].size(); i++) {
      int xx = e[x][y][i].first, yy = e[x][y][i].second;
      if (!col[xx][yy]) dfs(xx, yy, 5 - cl);
    }
  }
  int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) cin >> c[i] + 1;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++)
        if (c[i][j] == 'X') {
          vector<pair<int, int> > cur;
          for (int k = 0; k <= 3; k++) {
            if (c[i + t[k][0]][j + t[k][1]] == '.')
              cur.push_back(make_pair(i + t[k][0], j + t[k][1]));
          }
          if (cur.size() & 1) {
            puts("NO");
            return 0;
          }
          if (!cur.size()) continue;
          cur.push_back(cur[0]);
          for (int k = 1; k < cur.size(); k++) {
            e[cur[k].first][cur[k].second].push_back(cur[k - 1]);
            e[cur[k - 1].first][cur[k - 1].second].push_back(cur[k]);
          }
        }
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        if (c[i][j] == '.' && !col[i][j]) {
          dfs(i, j, 1);
        }
      }
    puts("YES");
    for (int i = 1; i <= n; i++, puts(""))
      for (int j = 1; j <= m; j++)
        if (c[i][j] == 'X') {
          for (int k = 0; k <= 3; k++) {
            if (c[i + t[k][0]][j + t[k][1]] == '.')
              col[i][j] += col[i + t[k][0]][j + t[k][1]];
          }
          printf("%d ", col[i][j]);
        } else
          printf("%d ", col[i][j]);
    return 0;
  }
  