67 lines
1.8 KiB
Markdown
67 lines
1.8 KiB
Markdown
|
---
|
||
|
|
title: Lee's Algorithm
|
||
|
|
---
|
||
|
|
|
||
|
|
## Lee's Algorithm
|
||
|
|
|
||
|
|
The Lee algorithm is one possible solution for maze routing problems. It always gives an optimal solution, if one exists, but is
|
||
|
|
slow and requires large memory for dense layout.
|
||
|
|
|
||
|
|
### Understanding how it works
|
||
|
|
|
||
|
|
The algorithm is a `breadth-first` based algorithm that uses `queues` to store the steps. It usually uses the following steps:
|
||
|
|
|
||
|
|
1. Choose a starting point and add it to the queue.
|
||
|
|
2. Add the valid neighboring cells to the queue.
|
||
|
|
3. Remove the position you are on from the queue and continue to the next element.
|
||
|
|
4. Repeat steps 2 and 3 until the queue is empty.
|
||
|
|
|
||
|
|
### Implementation
|
||
|
|
|
||
|
|
C++ has the queue already implemented in the `<queue>` library, but if you are using something else you are welcome to implement
|
||
|
|
your own version of queue.
|
||
|
|
|
||
|
|
C++ code:
|
||
|
|
|
||
|
|
```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();
|
||
|
|
|
||
|
|
}
|
||
|
|
|
||
|
|
|
||
|
|
}
|
||
|
|
|
||
|
|
```
|
||
|
|
|
||
|
|
|