0a1eeea424
* fix: replace sh with shell fix replace terminal with shell fix replace node with js fix replace output with shell fix replace cs with csharp fix replace c++ with cpp fix replace c# with csharp fix replace javasctipt with js fix replace syntax with js fix replace unix with shell fix replace linux with shell fix replace java 8 with java fix replace swift4 with swift fix replace react.js with jsx fix replace javascriot with js fix replace javacsript with js fix replace c++ - with cpp fix: corrected various typos fix: replace Algorithm with nothing fix: replace xaml with xml fix: replace solidity with nothing fix: replace c++ with cpp fix: replace txt with shell fix: replace code with json and css fix: replace console with shell
1.8 KiB
1.8 KiB
title
| 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:
- Choose a starting point and add it to the queue.
- Add the valid neighboring cells to the queue.
- Remove the position you are on from the queue and continue to the next element.
- 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:
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();
}
}