我想制作一个程序,通过回溯来查找迷宫中从右上角到左下角的路径。输入数字为n和m,它们是矩形迷宫和迷宫的尺寸,字符“。”。表示您可以通过的图块,字符“ x”表示您无法通过的图块。我已经编写了代码,它相当简单,但是什么都没有显示,而应该显示“ da”(在塞尔维亚“是”上)和“ ne”(在塞尔维亚“否”上)。
#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;
}
样本输入:
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..
示例输出:da
答案 0 :(得分:1)
问题在于,如果您从(0, 0) -> (1, 0)
开始,然后在(1, 0)
,您可以再次回到(0, 0)
,这将永远循环。为了避免这种情况,我创建了一个visited
数组,如果已经访问了单元格true
,则其值为(x, y)
,否则为false
。
我已经用///////////// change here /////////////
评论标记了更改的地方
#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;
}
这是我对其进行测试的链接:https://ideone.com/vVqAjF