61 lines
		
	
	
		
			1.8 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
		
		
			
		
	
	
			61 lines
		
	
	
		
			1.8 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
|   | --- | |||
|  | title: Lee's Algorithm | |||
|  | localeTitle: 李的算法 | |||
|  | --- | |||
|  | ## 李的算法
 | |||
|  | 
 | |||
|  | Lee算法是迷宫路由问题的一种可能解决方案。它始终提供最佳解决方案,如果存在,但是 缓慢并需要大内存以进行密集布局。 | |||
|  | 
 | |||
|  | ### 了解它是如何工作的
 | |||
|  | 
 | |||
|  | 该算法是基于`breadth-first`算法,它使用`queues`来存储步骤。它通常使用以下步骤: | |||
|  | 
 | |||
|  | 1.  选择一个起点并将其添加到队列中。 | |||
|  | 2.  将有效的相邻单元添加到队列中。 | |||
|  | 3.  从队列中删除您所在的位置并继续下一个元素。 | |||
|  | 4.  重复步骤2和3,直到队列为空。 | |||
|  | 
 | |||
|  | ### 履行
 | |||
|  | 
 | |||
|  | C ++已经在`<queue>`库中实现了`<queue>` ,但如果您正在使用其他东西,欢迎您实现 你自己的队列版本。 | |||
|  | 
 | |||
|  | C ++代码: | |||
|  | 
 | |||
|  | ```c++ | |||
|  | int dl[] = {-1, 0, 1, 0}; // these arrays will help you travel in the 4 directions more easily  | |||
|  |  int dc[] = {0, 1, 0, -1};  | |||
|  |   | |||
|  |  queue<int> X, Y; // the queues used to get the positions in the matrix  | |||
|  |   | |||
|  |  X.push(start_x); //initialize the queues with the start position  | |||
|  |  Y.push(start_y);  | |||
|  |   | |||
|  |  void lee()  | |||
|  |  {  | |||
|  |   int x, y, xx, yy;  | |||
|  |   while(!X.empty()) // while there are still positions in the queue  | |||
|  |   {  | |||
|  |     x = X.front(); // set the current position  | |||
|  |     y = Y.front();  | |||
|  |     for(int i = 0; i < 4; i++)  | |||
|  |     {  | |||
|  |       xx = x + dl[i]; // travel in an adiacent cell from the current position  | |||
|  |       yy = y + dc[i];  | |||
|  |       if('position is valid') //here you should insert whatever conditions should apply for your position (xx, yy)  | |||
|  |       {  | |||
|  |           X.push(xx); // add the position to the queue  | |||
|  |           Y.push(yy);  | |||
|  |           mat[xx][yy] = -1; // you usually mark that you have been to this position in the matrix  | |||
|  |       }  | |||
|  |   | |||
|  |     }  | |||
|  |   | |||
|  |     X.pop(); // eliminate the first position, as you have no more use for it  | |||
|  |     Y.pop();  | |||
|  |   | |||
|  |   }  | |||
|  |   | |||
|  |   | |||
|  |  }  | |||
|  | 
 | |||
|  | ``` |