Solved

Posted on 2003-11-19

hi! all

i'm writng a code for image compression in which i have to read the data from file and store it in a matrix form. Second step involves sorting of data....data can be sort but i've to store their positions not the values in an array (acc. to the sorted values).

Suppose 64(3rd row & 5th col) is the max value in the matrix...then i've to store 3,5 position of 64 . Now problem is how do i store that. i think it's pretty simple but i'm not bale to get it.

This is done so that value's original location is not lost...since it's an image & after whatever operation we need to send that vlue back to its original position so as to get the original image back.

What kind of structure do i need to use.

Please if someone cud help me solving this prob.

thanks in advance

i'm writng a code for image compression in which i have to read the data from file and store it in a matrix form. Second step involves sorting of data....data can be sort but i've to store their positions not the values in an array (acc. to the sorted values).

Suppose 64(3rd row & 5th col) is the max value in the matrix...then i've to store 3,5 position of 64 . Now problem is how do i store that. i think it's pretty simple but i'm not bale to get it.

This is done so that value's original location is not lost...since it's an image & after whatever operation we need to send that vlue back to its original position so as to get the original image back.

What kind of structure do i need to use.

Please if someone cud help me solving this prob.

thanks in advance

4 Comments

int* sortArray(int* array, int size)

{

// check for minimal two items

if (size <= 1)

return NULL;

// create integer array for the sorted indices

int* idxArr = new int[size];

int nn = 0; // nn is the number of already sorted items

int n1 = 0; // n1 is the lower range index when searching binary

int n2 = 0; // n2 is the upper range index when searching binary

int n = 0; // n is the middle of the binary range

int i; // loop counter

// loop all items and build a sorted index array

for (i = 0; i < size; i++)

{

// i1 is the new item which index must be inserted to the index array

int i1 = array[i];

// init the binary range to the current size of the already sorted index array

n1 = 0;

n2 = nn-1;

// binary search

while (n1 <= n2)

{

// compute the middle of the current range for dividing the current range into two parts

n = (n1 + n2 + 1)/2;

// get the item to compare

int i2 = array[idxArr[n]];

// compare both items

// if the left item is less we make the lower part current

if (i1 < i2)

n2 = n - 1;

// if the left item is greater or equal we make the upper part current

else

n1 = n + 1;

}

// at end of the previous loop n1 is the new position where the current item index (i)

// has to be inserted to get the index array sorted

// if this position is not outside the current size (that means we have not to change the order)

// we move all indices on the right side for one position using memmove

if (n1 < nn)

{

memmove(&idxArr[n1+1], &idxArr[n1], (nn-n1)*sizeof(int));

}

idxArr[n1] = i;

nn++;

}

return idxArr;

}

#define MAXX 100

#define MAXY 100

int main(int argc, char* argv[])

{

int matrix[MAXY][MAXX] = { 0 };

// read matrix from somewhere

if (readMatrix(matrix, ...))

{

int* array = (int*)matrix;

int size = MAXX*MAXY;

int* idxArr = sortArray(array, size);

// now you may store the index array to a file

// it is as big as the original image file (if the values to sort are integers)

writeSortedIndex(idxArr, ...);

// if you need the coordinates e. g. of the biggest value you may calculate

int idx = idxArr[size-1];

int x = idx % MAXX;

int y = idx / MAXX;

// then, you got the max value by

int maxInt = matrix[x][y];

}

}

I assumed that your values are integers.

Hope, that helps

Alex

You can use an array of structures. Suppose you have M rows and N columns, then

struct

{

int x;

int y;

}array[M*N];

Thus this array then can be accessed easily as array[0].x or array[0].y which will give x and y of first element.

Thus you can store the x and y values easily and access them.

Dhyanesh

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