DFS maze problem

The title roughly describes:
The maze consists of n rows and m columns of cells (n, m are less than or equal to 50)
Each cell is either an open space or an obstacle
Find the shortest path length from the start to the end
  1  / * 
  2  * @Descripttion: 
   3  * @version: 
   4  * @Author: ZKYAAA
   5  * @Date: 2020-04-20 22:25:27
   6  * @LastEditors: Please call me ZK 訕 啊啊 啊
   7  * @ LastEditTime: 2020-04-20 23:42:48
   8  
  9  * Problem description:
 10  * The maze is composed of n rows and m columns of cells (n, m are less than or equal to 50)
 11  * Each cell is either open or It is an obstacle
 12  * Find the shortest path length from the start point to the end point
 13  
14  5 4
 15  1 1 2 1 
 16  1 1 1 1 1
 17  1 1 2 1
 18  1 2 1 1
 19  1 1 1 2
 20  1 1 4 3
 21  * /
22  
23 #include <bits / stdc ++. H>
 24  using  namespace std;
 25  const  int MAXN = 100 ;
 26  int p, q, Min = 1e9, n, m;
 27  int a [MAXN] [MAXN];       // 1 Means open space, 2 means obstacle 
28  int vis [MAXN] [MAXN];     // 0 means no visit, 1 means visit 
29  
30  int dir [ 4 ] [ 2 ] = {
 31      { 0 , 1 },
 32      { 1 , 0 },
33      { 0 , -1 },
 34      { -1 , 0 },
 35  };
 36  
37  // int dirx [4] = {0,1,0, -1};      // The four directions are bottom left and top right
 38  // int diry [4] = {1,0, -1,0}; 
39  
40  void dfs ( int x, int y, int step) {
 41      if (x == p && y == q) {
 42          if (step < Min ) {
 43              Min = step;
 44          }
 45         return ;          // backtrack 
46      }
 47      // clockwise lower right upper left trial
 48      // optimized code 
49      for ( int k = 0 ; k < 4 ; k ++ ) {
 50          // int dx = x + dirx [k] ;
 51          // int dy = y + diry [k]; 
52          int dx = x + dir [k] [ 0 ];
 53          int dy = y + dir [k] [ 1 ];
 54          if (a [dx] [ dy] == 1 && vis [dx] [dy] == 0 ) {
 55              vis [dx] [dy] =1 ;
 56              dfs (dx, dy, step + 1 );
 57              vis [dx] [dy] = 0 ;
 58          }
 59      }
 60      / * 
61      // right
 62      if (a [x] [y + 1] == 1 && vis [x] [y + 1] == 0) // Determine whether the right side of (x, y) is visited and is not an obstacle
 63      {
 64          vis [x] [y + 1] = 1;
 65          dfs (x, y + 1, step + 1);
 66          vis [x] [y + 1] = 0; // back to this point after dfs is executed, mark unvisited, continue to find step
 67      }
 68      // Next
 69      if (a [x +1] [y] == 1 && vis [x + 1] [y] == 0) // Determine whether the bottom of (x, y) is visited and is not an obstacle
70      {
 71          vis [x + 1] [y] = 1;
 72          dfs (x + 1, y, step + 1);
 73          vis [x + 1] [y] = 0; // dfs back to this after execution Point, mark not visited, continue to find step
 74      }
 75      // left
 76      if (a [x-1] [y] == 1 && vis [x-1] [y] == 0) // judgment (x, y) Is the left side of the site visited and not an obstacle
 77      {
 78          vis [x-1] [y] = 1;
 79          dfs (x-1, y, step + 1);
 80          vis [x-1] [y] = 0 ; // dfs traces back to this point after execution, marks not visited, continue to find step
 81      }
 82      // up
 83      if (a [x] [y-1] == 1 && vis [x] [y-1] == 0 ) // determine whether (x, y) and the top is not an obstacle to access
 84     {
 85          vis [x] [y-1] = 1;
 86          dfs (x, y-1, step + 1);
 87          vis [x] [y-1] = 0; // dfs back to this point after execution , Mark not visited, continue to find step
 88      }
 89     * / 
90      return ;                                  // back 
91  }
 92  
93  int main () {
 94      int startx, starty;
 95      memset (vis, 0 , sizeof (vis));
 96      cin >> m >> n;
 97      // At the beginning, it was written as n first and then m to create a labyrinth. It has been running with error 5 
98      for (int i=1;i<=m;i++)
 99         for(int j=1;j<=n;j++)
100             cin>>a[i][j];
101     cin>>startx>>starty>>p>>q;
102     vis[startx][starty]=1;
103     dfs(startx,starty,0);
104     cout<<Min;
105     return 0;
106 }
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