给定一个N * N二元迷宫，其中0表示可以访问该位置，而1表示没有密钥就无法访问该位置，任务是找到是否可以从顶部访问右下角的单元格- 沿途只有一个键的左单元格。如果可能，则打印“是”，否则打印“否” 。
例子：
Input: maze[][] = {
{0, 0, 1},
{1, 0, 1},
{1, 1, 0}}
Output: Yes
方法：这个问题可以使用递归来解决，对于每一个可能的移动，如果当前单元格为0则不改变键的状态检查它是否是目的地否则向前移动。如果当前单元格为1，则必须使用密钥，现在为了进一步移动，密钥将设置为false，即它永远不会在同一路径上再次使用。如果任何路径到达目的地，则打印Yes否则打印No 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Recursive function to check whether there is
// a path from the top left cell to the
// bottom right cell of the maze
bool findPath(vector> maze,
int xpos, int ypos, bool key)
{
// Check whether the current cell is
// within the maze
if (xpos < 0 || xpos >= maze.size() ||
ypos < 0 || ypos >= maze.size())
return false;
// If key is required to move further
if (maze[xpos][ypos] == '1')
{
// If the key hasn't been used before
if (key == true)
// If current cell is the destination
if (xpos == maze.size() - 1 &&
ypos == maze.size() - 1)
return true;
// Either go down or right
return findPath(maze, xpos + 1,
ypos, false) ||
findPath(maze, xpos,
ypos + 1, false);
// Key has been used before
return false;
}
// If current cell is the destination
if (xpos == maze.size() - 1 &&
ypos == maze.size() - 1)
return true;
// Either go down or right
return findPath(maze, xpos + 1,
ypos, key) ||
findPath(maze, xpos,
ypos + 1, key);
}
bool mazeProb(vector> maze,
int xpos, int ypos)
{
bool key = true;
if (findPath(maze, xpos, ypos, key))
return true;
return false;
}
// Driver code
int main()
{
vector> maze = { { '0', '0', '1' },
{ '1', '0', '1' },
{ '1', '1', '0' } };
int n = maze.size();
// If there is a path from the cell (0, 0)
if (mazeProb(maze, 0, 0))
cout << "Yes";
else
cout << "No";
}
// This code is contributed by grand_master Java
// Java implementation of the approach
import java.io.*;
import java.util.ArrayList;
class GFG{
// Recursive function to check whether there
// is a path from the top left cell to the
// bottom right cell of the maze
static boolean findPath(
ArrayList> maze,
int xpos, int ypos, boolean key)
{
// Check whether the current cell is
// within the maze
if (xpos < 0 || xpos >= maze.size() ||
ypos < 0 || ypos >= maze.size())
return false;
// If key is required to move further
if (maze.get(xpos).get(ypos) == '1')
{
// If the key hasn't been used before
if (key == true)
// If current cell is the destination
if (xpos == maze.size() - 1 &&
ypos == maze.size() - 1)
return true;
// Either go down or right
return findPath(maze, xpos + 1, ypos, false) ||
findPath(maze, xpos, ypos + 1, false);
}
// If current cell is the destination
if (xpos == maze.size() - 1 &&
ypos == maze.size() - 1)
return true;
// Either go down or right
return findPath(maze, xpos + 1, ypos, key) ||
findPath(maze, xpos, ypos + 1, key);
}
static boolean mazeProb(
ArrayList> maze,
int xpos, int ypos)
{
boolean key = true;
if (findPath(maze, xpos, ypos, key))
return true;
return false;
}
// Driver code
public static void main(String[] args)
{
int size = 3;
ArrayList> maze =
new ArrayList>(size);
for(int i = 0; i < size; i++)
{
maze.add(new ArrayList());
}
// We are making these
//{ { '0', '0', '1' },
// { '1', '0', '1' },
// { '1', '1', '0' } };
maze.get(0).add(0);
maze.get(0).add(0);
maze.get(0).add(1);
maze.get(1).add(1);
maze.get(1).add(0);
maze.get(1).add(1);
maze.get(2).add(1);
maze.get(2).add(1);
maze.get(2).add(0);
// If there is a path from the cell (0, 0)
if (mazeProb(maze, 0, 0))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by sujitmeshram Python3
# Python3 implementation of the approach
# Recursive function to check whether there is
# a path from the top left cell to the
# bottom right cell of the maze
def findPath(maze, xpos, ypos, key):
# Check whether the current cell is
# within the maze
if xpos < 0 or xpos >= len(maze) or ypos < 0 \
or ypos >= len(maze):
return False
# If key is required to move further
if maze[xpos][ypos] == '1':
# If the key hasn't been used before
if key == True:
# If current cell is the destination
if xpos == len(maze)-1 and ypos == len(maze)-1:
return True
# Either go down or right
return findPath(maze, xpos + 1, ypos, False) or \
findPath(maze, xpos, ypos + 1, False)
# Key has been used before
return False
# If current cell is the destination
if xpos == len(maze)-1 and ypos == len(maze)-1:
return True
# Either go down or right
return findPath(maze, xpos + 1, ypos, key) or \
findPath(maze, xpos, ypos + 1, key)
def mazeProb(maze, xpos, ypos):
key = True
if findPath(maze, xpos, ypos, key):
return True
return False
# Driver code
if __name__ == "__main__":
maze = [['0', '0', '1'],
['1', '0', '1'],
['1', '1', '0']]
n = len(maze)
# If there is a path from the cell (0, 0)
if mazeProb(maze, 0, 0):
print("Yes")
else:
print("No")C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
// Recursive function to check whether there
// is a path from the top left cell to the
// bottom right cell of the maze
static bool findPath(List> maze,
int xpos, int ypos, bool key)
{
// Check whether the current cell is
// within the maze
if (xpos < 0 || xpos >= maze.Count ||
ypos < 0 || ypos >= maze.Count)
return false;
// If key is required to move further
if (maze[xpos][ypos] == '1')
{
// If the key hasn't been used before
if (key == true)
// If current cell is the destination
if (xpos == maze.Count - 1 &&
ypos == maze.Count - 1)
return true;
// Either go down or right
return findPath(maze, xpos + 1, ypos, false) ||
findPath(maze, xpos, ypos + 1, false);
}
// If current cell is the destination
if (xpos == maze.Count - 1 &&
ypos == maze.Count - 1)
return true;
// Either go down or right
return findPath(maze, xpos + 1, ypos, key) ||
findPath(maze, xpos, ypos + 1, key);
}
static bool mazeProb(List> maze,
int xpos, int ypos)
{
bool key = true;
if (findPath(maze, xpos, ypos, key))
return true;
return false;
}
// Driver code
public static void Main(String[] args)
{
int size = 3;
List> maze =
new List>(size);
for(int i = 0; i < size; i++)
{
maze.Add(new List());
}
// We are making these
//{ { '0', '0', '1' },
// { '1', '0', '1' },
// { '1', '1', '0' } };
maze[0].Add(0);
maze[0].Add(0);
maze[0].Add(1);
maze[1].Add(1);
maze[1].Add(0);
maze[1].Add(1);
maze[2].Add(1);
maze[2].Add(1);
maze[2].Add(0);
// If there is a path from the cell (0, 0)
if (mazeProb(maze, 0, 0))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by gauravrajput1 Javascript
输出：
Yes时间复杂度： O(2 N )
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