[英]Finding a path in a maze via backtracking
I want to make a program to find if a path exists from upper right corner to down left corner in a maze via backtracking.我想编写一个程序,通过回溯查找迷宫中从右上角到左下角的路径是否存在。 The input numbers are n and m which are the dimensions of rectangular maze and a maze, character '.'输入数字是 n 和 m,它们是矩形迷宫和迷宫的维度,字符 '.' means a tile which you can go through and character 'x' means a tile which you cant go through.表示您可以通过 go 的图块,字符“x”表示您无法通过的图块 go。 I have wrote the code, its fairly simple but nothing gets displayed whilst it should display "da" (on Serbian "yes") and "ne" (on Serbian "no").我已经编写了代码,它相当简单但没有显示任何东西,而它应该显示“da”(塞尔维亚语“是”)和“ne”(塞尔维亚语“否”)。
#include <bits/stdc++.h>
using namespace std;
bool maze[20][20]; //defined maze of maximum size 20x20
//checking if a position is viable for moving through
bool Safe(int n, int m, int x, int y)
{
if(x >= 0 && x < n && y >= 0 && y < m)
{
if(maze[x][y] == 1) return true;
}
return false;
}
bool Utility(int n, int m, int x, int y) //main utility function
{
if(x == n - 1 && y == m - 1 && maze[x][y] == 1) // base case, end of maze
{
return true;
}
if(Safe(n, m, x, y))
{
if(Safe(n, m, x + 1, y)) // checking if it is viable to move down
{
if(Utility(n, m, x + 1, y))
{
return true;
}
}
if(Safe(n, m, x, y + 1))
{
if(Utility(n, m, x, y + 1)) // checking if it is viable to move right
{
return true;
}
}
if(Safe(n, m, x - 1, y))
{
if(Utility(n, m, x - 1, y)) // checking if it is viable to move up
{
return true;
}
}
if(Safe(n, m, x, y - 1))
{
if(Utility(n, m, x, y - 1)) // checking if it is viable to move left
{
return true;
}
}
}
return false; // returning false
}
int main()
{
int n, m;
cin >> n >> m; // input dimensions of the maze
for(int i = 0; i < n; i++) // input maze
{
for(int j = 0; j < m; j++)
{
char c;
cin >> c;
if(c == '.') //character '.' means a tile which you can go through
{
maze[i][j] = 1;
}
else //character 'x' means a tile which you cannot go through
{
maze[i][j] = 0;
}
}
}
if(Utility(n, m, 0, 0)) //printing yes or no
{
cout << "da";
}
else
{
cout << "ne";
}
return 0;
}
Sample Input:示例输入:
8 8
.x.....x
.x.x.x.x
.x.x.x.x
.x.x.x.x
.x.x.x.x
.x.x.x.x
.x.x.x.x
...x.x..
Sample output: da
样本 output: da
The problem was that, say if you go from (0, 0) -> (1, 0)
, then at (1, 0)
you can again go back to (0, 0)
and this would loop forever.问题是,如果你 go 从(0, 0) -> (1, 0)
,然后在(1, 0)
你可以再次 go 回到(0, 0)
并且这将永远循环。 To avoid that, I created a visited
array which will have value true
if cell (x, y)
is already visited, else false
.为避免这种情况,我创建了一个visited
数组,如果单元格(x, y)
已被访问,则该数组的值为true
,否则为false
。
I have marked where I made the changes with ///////////// change here /////////////
comment我已经用///////////// change here /////////////
注释标记了我进行更改的位置
#include <bits/stdc++.h>
using namespace std;
bool maze[20][20]; //defined maze of maximum size 20x20
///////////// change here /////////////
bool visited[20][20];
bool Safe(int n, int m, int x, int y) //checking if a position is viable for moving through
{
if(x >= 0 && x < n && y >= 0 && y < m)
{
if(maze[x][y] == 1) return true;
}
return false;
}
bool Utility(int n, int m, int x, int y) //main utility function
{
if(x == n - 1 && y == m - 1 && maze[x][y] == 1) // base case, end of maze
{
return true;
}
///////////// change here /////////////
if(!visited[x][y] && Safe(n, m, x, y))
{
///////////// change here /////////////
visited[x][y] = true;
if(Safe(n, m, x + 1, y)) // checking if it is viable to move down
{
if(Utility(n, m, x + 1, y))
{
return true;
}
}
if(Safe(n, m, x, y + 1))
{
if(Utility(n, m, x, y + 1)) // checking if it is viable to move right
{
return true;
}
}
if(Safe(n, m, x - 1, y))
{
if(Utility(n, m, x - 1, y)) // checking if it is viable to move up
{
return true;
}
}
if(Safe(n, m, x, y - 1))
{
if(Utility(n, m, x, y - 1)) // checking if it is viable to move left
{
return true;
}
}
}
return false; // returning false
}
int main()
{
int n, m;
cin >> n >> m; // input dimensions of the maze
for(int i = 0; i < n; i++) // input maze
{
for(int j = 0; j < m; j++)
{
char c;
cin >> c;
if(c == '.') //character '.' means a tile which you can go through
{
maze[i][j] = true;
}
else //character 'x' means a tile which you cannot go through
{
maze[i][j] = false;
}
///////////// change here /////////////
visited[i][j] = false;
}
}
if(Utility(n, m, 0, 0)) //printing yes or no
{
cout << "da";
}
else
{
cout << "ne";
}
return 0;
}
Here's the link where I tested it: https://ideone.com/vVqAjF这是我测试它的链接: https://ideone.com/vVqAjF
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