Rat in maze problems, how to display coordinates where rat stepped in random generated matrix?

Question

I started coding and searched a lot codes on internet but couldn't find the solution for my problem. I'm gonna paste there my current code. I already coded that if there is perfect path for rat or no (rat can only step on number one, "0" and "2" are walls.) But I cant display the coordinates of each "number one". For an example:

1,0,1,
1,2,2,
1,1,1,

Then the path is, (0,0) -> (1,0) -> (2,0) -> (1,2) -> (2,2)

if the matrix is 1,0,2, 0,2,1, 1,0,1, Then no path for the rat. Please edit my code and paste it there please or if you have already solved program for my problem please share it with me... I'm really new in c++.

// CPP program to solve Rat in a maze
// problem with backtracking using stack

#include <cstring>
#include <iostream>
#include <stack>

using namespace std;

#define N 4
#define M 5
void mtxkeres(int (&mat)[N][M]);
class node {
public:
    int x, y;
    int dir;

    node(int i, int j)
    {
        x = i;
        y = j;

        // Initially direction
        // set to 0
        dir = 0;
    }
};

// maze of n*m matrix
int n = N, m = M;

// Coordinates of food
int fx, fy;
bool visited[N][M];

bool isReachable(int maze[N][M])
{
    // Initially starting at (0, 0).
    int i = 0, j = 0;

    stack<node> s;

    node temp(i, j);

    s.push(temp);

    while (!s.empty()) {

        // Pop the top node and move to the
        // left, right, top, down or retract
        // back according the value of node's
        // dir variable.
        temp = s.top();
        int d = temp.dir;
        i = temp.x, j = temp.y;

        // Increment the direction and
        // push the node in the stack again.
        temp.dir++;
        s.pop();
        s.push(temp);

        // If we reach the Food coordinates
        // return true
        if (i == fx and j == fy) {
            return true;
        }

        // Checking the Up direction.
        if (d == 0) {
            if (i - 1 >= 0 and maze[i - 1][j] and
                                    visited[i - 1][j]) {
                node temp1(i - 1, j);
                visited[i - 1][j] = false;
                s.push(temp1);
            }
        }

        // Checking the left direction
        else if (d == 1) {
            if (j - 1 >= 0 and maze[i][j - 1] and
                                    visited[i][j - 1]) {
                node temp1(i, j - 1);
                visited[i][j - 1] = false;
                s.push(temp1);
            }
        }

        // Checking the down direction
        else if (d == 2) {
            if (i + 1 < n and maze[i + 1][j] and
                                    visited[i + 1][j]) {
                node temp1(i + 1, j);
                visited[i + 1][j] = false;
                s.push(temp1);
            }
        }
        // Checking the right direction
        else if (d == 3) {
            if (j + 1 < m and maze[i][j + 1] and
                                    visited[i][j + 1]) {
                node temp1(i, j + 1);
                visited[i][j + 1] = false;
                s.push(temp1);
            }
        }

        // If none of the direction can take
        // the rat to the Food, retract back
        // to the path where the rat came from.
        else {
            visited[temp.x][temp.y] = true;
            s.pop();
        }
    }

    // If the stack is empty and
    // no path is found return false.
    return false;
}

// Driver code
int main()
{
    // Initially setting the visited
    // array to true (unvisited)
    memset(visited, true, sizeof(visited));

    // Maze matrix
    int maze[N][M] = {
        { 1, 0, 1, 1, 0 },
        { 1, 1, 1, 0, 1 },
        { 0, 1, 0, 1, 1 },
        { 1, 1, 1, 1, 1 }
    };

    // Food coordinates
    fx = 2;
    fy = 3;

    if (isReachable(maze)) {
        cout << "Path Found!" << '\n';
    }
    else
        cout << "No Path Found!" << '\n';

    return 0;
}