I'm starting to learn how to implement divide and conquer algorithms, but I'm having some trouble with this exercise.
I have written an algorithm but unfortunately it returns a 0 value. I need to count how many times a number finds in the matrix using D&C.
My idea is to divide the matrix in 4 sub-matrices till all coordinates are equal, it means that there is an element (could be my number or not).
My abbreviations:
lr - left row | starting with
lc- left column | left lower corner
rr - right row | starting with
rc- right column | right upper corner
When I compile, it runs well but returns 0 (because of my stop statement I think). So, my error is in solve() function.
My code:
#include<fstream>
#include<conio.h>
#include<iostream>
using namespace std;
ifstream fin("in.txt");
#define debug cerr<<"OK";
int n,a[100][100];
int solve(int lr,int lc, int rr,int rc,int x,int &contor)
{
int cnt=0;
if(lr < rr || lc > rc)
{
if(cnt<=0)
return 0;
return cnt;
}
if(lr == lc && rr == rc)
if(a[lr][lc] == x)
cnt++;
else;
else
{
int l=(lr+rr)/2;
int c=(lc+rc)/2;
if(l && c)
{
int cnt1=solve(lr-l,lc,rr,rc-c,x,contor); // coordinates for upper left square
int cnt2=solve(lr-l,lc+c,rr,rc,x,contor); // -||- for upper right square
int cnt3=solve(lr,lc,rr+l,rc-c,x,contor); // -||- for low left square
int cnt4=solve(lr,lc+c,rr,rc-c,x,contor); // -||- for low right square
contor=cnt1+cnt2+cnt3+cnt4;
}
}
}
int main()
{
fin>>n;
int i,j;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
fin>>a[i][j];
int contor=0;
solve(n,1,1,n,3,contor); // i'm searching for 3
cout<<contor;
_getch();
fin.close();
return 0;
}
Maybe you can tell me where the problem is or directly solve it, it would be faster and I can understand easier.
Input:
4
1 3 3 4
5 6 7 8
1 2 3 4
4 3 2 1
Here is the corrected source code. I have added some comments so you can follow it. Furthermore, I switched to a dynamic array of the correct size.
#include<fstream>
#include<conio.h>
#include<iostream>
using namespace std;
int countOccurences(int** values, int min1, int min2, int max1, int max2, int searchValue)
{
if (min1 > max1 || min2 > max2)
return 0; //invalid area
if (min1 == max1 && min2 == max2)
{
//the current area is a single cell
if (values[min1][min2] == searchValue)
return 1; //we have found 1 occurence
else
return 0; //we have found nothing
}
else
{
//divide the current range
int center1 = (min1 + max1) / 2;
int center2 = (min2 + max2) / 2;
int occurencesInSubMatrices = 0;
//accumulate the occurences in the according variable
occurencesInSubMatrices += countOccurences(values, min1, min2, center1, center2, searchValue); // coordinates for upper left square
occurencesInSubMatrices += countOccurences(values, center1 + 1, min2, max1, center2, searchValue); // -||- for upper right square
occurencesInSubMatrices += countOccurences(values, min1, center2 + 1, center1, max2, searchValue); // -||- for low left square
occurencesInSubMatrices += countOccurences(values, center1 + 1, center2 + 1, max1, max2, searchValue); // -||- for low right square
return occurencesInSubMatrices;
}
}
int main()
{
ifstream fin("in.txt");
int n;
fin >> n;
//create a 2d array of appropriate size
int** values = new int*[n];
for (int i = 0; i < n; i++)
{
//allocate memory for the i-th row
values[i] = new int[n];
for (int j = 0; j < n; j++)
fin >> values[i][j];
}
int count = countOccurences(values, 0, 0, n - 1, n - 1, 3); // i'm searching for 3
cout << count;
_getch();
fin.close();
return 0;
}
Please be aware that this is a purely educational example. In the real world, nobody would count ocurrences like this.