TheAlgorithms-C-Plus-Plus/backtracking/rat_maze.cpp

/*
	A Maze is given as N*N binary matrix of blocks where source block is the upper
	left most block i.e., maze[0][0] and destination block is lower rightmost 
	block i.e., maze[N-1][N-1]. A rat starts from source and has to reach destination.
	The rat can move only in two directions: forward and down. In the maze matrix, 
	0 means the block is dead end and 1 means the block can be used in the path 
	from source to destination.
*/
#include <iostream>
#define size 4

using namespace std;

int solveMaze(int currposrow, int currposcol, int maze[size][size], int soln[size][size])
{
	if ((currposrow == size - 1) && (currposcol == size - 1))
	{
		soln[currposrow][currposcol] = 1;
		for (int i = 0; i < size; ++i)
		{
			for (int j = 0; j < size; ++j)
			{
				cout << soln[i][j];
			}
			cout << endl;
		}
		return 1;
	}
	else
	{
		soln[currposrow][currposcol] = 1;

		// if there exist a solution by moving one step ahead in a collumn
		if ((currposcol < size - 1) && maze[currposrow][currposcol + 1] == 1 && solveMaze(currposrow, currposcol + 1, maze, soln))
		{
			return 1;
		}

		// if there exists a solution by moving one step ahead in a row
		if ((currposrow < size - 1) && maze[currposrow + 1][currposcol] == 1 && solveMaze(currposrow + 1, currposcol, maze, soln))
		{
			return 1;
		}

		// the backtracking part
		soln[currposrow][currposcol] = 0;
		return 0;
	}
}

int main(int argc, char const *argv[])
{
	int maze[size][size] = {
		{1, 0, 1, 0},
		{1, 0, 1, 1},
		{1, 0, 0, 1},
		{1, 1, 1, 1}};

	int soln[size][size];

	for (int i = 0; i < size; ++i)
	{
		for (int j = 0; j < size; ++j)
		{
			soln[i][j] = 0;
		}
	}

	int currposrow = 0;
	int currposcol = 0;
	solveMaze(currposrow, currposcol, maze, soln);
	return 0;
}