改善迷宫搜索

时间:2013-04-19 10:15:26

标签: java recursion backtracking maze

我做了一个简单的迷宫搜索算法,盲目地进入细胞的每个方向 检查目标是否已找到。它可以找到目标,但解决方案仍然很糟糕,因为它在找到目标时不会终止其他递归调用 并且它绘制了它已经走过的所有路径,而不仅是通往目标的一条可能路径。我怎么能改善这个?

伪代码中的基本算法是这样的:

function searchMaze(start_point, end_point, maze) {

 // If current point is goal I would like to abort other recursive calls
 if(start_point.equals(end_point)) {
   pathFound = true;
   drawPathInArray(); 
    return;
  }
 else {
    // if current point is not inside the array
   if(start_point_not_within_array)
   return
   else {
     // if current point is a wall or a cell that has already been visited
     if(cell_is_wall || cell_is_visited)
       return;
      else {
           // mark cell as possible path
           markCellInPathArray("#");
           setCellVisited();
           searchMaze(left_from_start_point, end_point, maze);
           searchMaze(right_from_start_point, end_point, maze);
           searchMaze(above_start_point, end_point, maze);
           searchMaze(below_start_point, end_point, maze); 
      }
   }
  }
}

3 个答案:

答案 0 :(得分:1)

让函数返回一个布尔值。每当您找到目标的路径时,返回true,否则返回false。一旦从一个递归调用中获得返回值true,就返回。

所涉及的更改将是适当的返回语句,并将您的4个递归调用更改为:

if (searchMaze(left_from_start_point, end_point, maze))
  return true;
if (searchMaze(right_from_start_point, end_point, maze))
  return true;
if (searchMaze(above_start_point, end_point, maze))
  return true;
if (searchMaze(below_start_point, end_point, maze))
  return true;

另外,你最后不应该有setCellNotVisited()吗?

注意:我看到你已经有一个pathFound变量(可能是一个类变量)。在这种情况下,Marco的解决方案可能更受欢迎,但最好将其更改为返回值。

答案 1 :(得分:1)

您应该将标志添加到每个递归调用中:

pathFound = pathFound || searchMaze(left_from_start_point, end_point, maze);
pathFound = pathFound || searchMaze(right_from_start_point, end_point, maze);
pathFound = pathFound || searchMaze(above_start_point, end_point, maze);
pathFound = pathFound || searchMaze(below_start_point, end_point, maze);

如果pathFound为true,则忽略调用。

答案 2 :(得分:1)

在else块中,如果为

设置为true
1) cell_is_wall when wall is visited 
2) set start_point_not_within_arrary when it is not in the array

那么你的代码应该可行。其他递归调用将由您已经拥有的这些条件检查处理。