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;
}