not able to trace grey cell in maze - c++

Question

There is a matrix which contains white cells(represented as 1) , black cells(represented as 0) and only one gray cell(represented as 2), need to go from (0,0) to (N-1, N-1) in Array[N][N].
Constraints:
1) The path should cover only white cells and must go via grey cell (this grey cell can be anywhere in the array)
2) The node once visited cannot be visited again.
Below is the typical maze problem solution but this solution doesn't handle the specific case of traversing GREY cell...so can you please help me in modifying the below code to handle the specific case.
My problem is that I am not sure how to put a check for grey cell?
#include "stdafx.h"
#include "algorithm"
#include <iostream>
#include <fstream>
using namespace std;
#include<stdio.h>
// Maze size
#define N 4
bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N]);
/* A utility function to print solution matrix sol[N][N] */
void printSolution(int sol[N][N])
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
printf(" %d ", sol[i][j]);
printf("\n");
}
}
/* A utility function to check if x,y is valid index for N*N maze */
bool isSafe(int maze[N][N], int x, int y)
{
//solveMazeUtil() to solve the problem. It returns false if no path is possible,
//otherwise return true and prints the path in the form of 1s. Please note that
//there may be more than one solutions, this function prints one of the feasible
if(x >= 0 && x < N && y >= 0 && y < N && maze[x][y] == 1)
// if (x,y outside maze) return false
return true;
return false;
}
/* This function solves the Maze problem using Backtracking. It mainly uses
solutions.*/
bool solveMaze(int maze[N][N])
{
int sol[N][N] = { {0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 0}
};
if(solveMazeUtil(maze, 0, 0, sol) == false)
{
printf("Solution doesn't exist");
return false;
}
printSolution(sol);
return true;
}
/* A recursive utility function to solve Maze problem */
bool solveMazeUtil(int maze[N][N], int x, int y, int sol[N][N])
{
// if (x,y is goal) return true
if(x == N-1 && y == N-1)
{
sol[x][y] = 1;
return true;
}
// Check if maze[x][y] is valid
if(isSafe(maze, x, y) == true)
{
// mark x,y as part of solution path
sol[x][y] = 1;
/* Move forward in x direction */
if (solveMazeUtil(maze, x+1, y, sol) == true)
return true;
/* If x moving in x direction doesn't give solution then
Move down in y direction */
if (solveMazeUtil(maze, x, y+1, sol) == true)
return true;
/* If none of the above movements work then BACKTRACK:
unmark x,y as part of solution path */
sol[x][y] = 0;
return false;
}
return false;
}
// driver program to test above function
int main()
{
int maze[N][N] = { {1, 0, 0, 0},
{1, 1, 0, 1},
{0, 1, 0, 0},
{1, 1, 1, 1}
};
solveMaze(maze);
getchar();
return 0;
}
One solution what I am thinking is:
Produce all the possible paths (which traverse through either 1 or 2).
Then, find out which of the path is having 2 in it. and then print that path as the output.
But I don't think this will be the good approach...So, please let me know how to achieve my goal in a decent way.
Thanks

Answer 1

Since in your code you are only using two possible movements: down and right then this is a DAG. A DAG is suitable for a dynamic programming approach: each cell has two possibilities to get there, one is coming from above and the other is coming from left. Thus the minimum distance for a cell is:
cost[i][j] = min(cost[i][j-1],cost[i-1][j]) + 1
That is considering that the cost to do a movement is 1. If the cell is black you can give it an infinite cost, and you only need to find a path from P1(start) to P2(gray cell) and then a path from P2 to P3(goal).
For reconstructing the path, you can create another matrix of parents pi[N][N], if shortest path is coming from above then pi[i][j] = (i-1, j) if is coming from left pi[i][j] = (i, j-1) if is impossible to reach that cell pi[i][j] = null(whatever you want).

Answer 2

In general, my approach would be:
Generate a graph where each cell is a vertex, and connect with an edge vertices that represent adjacent white/grey cells in the maze.
Find the shortest path P1 between the the starting vertex (the one representing Array[0][0]) and the grey vertex (A* would be recommended).
Find the shortest path P2 between the the grey vertex and the end vertex (the one representing Array[N-1][N-1]).
P1 and P2 may intesect only once (as they represent the shortest paths), if they do intersect, from this point and on the will represent the same path. thus:
if P1 and P2 don't intersect, then P1 followed by P2 is the optimal solution.
if P1 and P2 do intersect, then delete the vertices in the intersected part, peform actions 2 and 3 again to find new shortest paths P3 and P4 respectively. the optimal solution is the minimum between P1 followed by P4 and P3 followed by P2.