Maze with exact moves available to finish

[schobelix]

Hi, I'd like to ask for help. I neeed to print maze made of '.' and '#' where '#' is wall and '.' is just place I can move at.

Maze has to be generated with exact rows and columns depending on the users input, but I've already done that with one-dimensional array (but I'm still not sure if it woud work or I will have to do that with two-dimensional array).

Start is at the upper left-hand corner and the end is at the lower right-hand corner.
Variable k has to be at the interval I've written (lowk and highk)

I need to find and write algorithm that will successfully lead you from start to finish at the exact moves (variable k)=> path will be printed from '.' and that will write '#' (walls) so that number of moves is equal to variable k.
This number of moves cannot exceed or be lower than k.

#include <iostream>
#include <math.h>

using namespace std;

int main()
{
    int wardens, rows, columns, k, lowk, highk, columner;
    columner=0;
    cout << "Enter number of wardens (from 1 to 1000)" << endl;
    cin >> wardens;
    for (int i=1;i<=wardens;i++)
    {
        lowk=0;
        highk=0;
        cout << "Dimensions of " << i << ". warden:" << endl;
        cout << "Enter number of rows (from 1 to 75)(1 <= rows <= 75)" << endl; //rows
        cin >> rows;
        cout << "Enter number of columns (from 1 to 75)(1 <= columns <= 75)" << endl; //columns
        cin >> columns;
        lowk=columns+rows-1;
        highk=2*(((rows-1)/4)*((columns-1)%4)+((columns-1)/4)*(rows-1))+rows+columns-1;
        cout << "Enter k (from " << lowk << " to "<< highk << ")(" << lowk << " <= k <= " << highk << ")" << endl; // distance/lenght of shortest way/number of moves
        cin >> k;
        char array[rows*n];
        for (int r=0;r<(rows*columns);r++)
        {
            array[r]='.';
            if (array[r]==array[0])
                array[r]='.'; //first position will be always '.'
            else if (array[r]==array[(rows*columns)-1])
                array[r]='.'; //last position will be always '.'

            cout << array [r] << " ";
            columner++;
            if (columner%columns==0) // to have X columns at Y rows
                cout << endl;
        }
    }

    return 0;
}

if in input:
wardens=(any)
rows=5
columns=6
k=12
output:

......
......
......
......
......

desired output example:

......
..##..
.#....
.##.##
#.....

[dhayden]

I'd use a 2D array for sure. Let the compiler do the arithmetic for you.

Line 25: n is undefined so this doesn't compile. When posting code and results, please be sure that the code generates the results.

Lines 26-31:
The first time through the loop, r==0, so line 28 sets array[0]='.'. Line 29 is true so line 30 also sets array[0]='.'.

The second time through the loop, r==1. Line 28 sets array[1]='.'. Line 29 compares array[1] to array[0] and since they're both '.', it's true and line 30 sets array[1]='.' again.

The pattern repeats. For each value of r, line 28 sets array[r]='.' and the comparison at line 29 is true.

You desired output contains '#' characters, but I don't see anything in the code that would put '#' in any location.

[schobelix]

there's supposed to be variable columns instead of n, im sorry for that.

I know that this code will print only '.' but that's just because I dont know how write the algorithm that will print number of '#' correctly so that shortest way made of '.' will match the value of variable k. Im trying to figure it out how could it work.