Number of closed rectangles in a matrix

Given a matrix containing only 1 and 0, find the number of closed rectangles. A closed rectangle is represented by at least one 0 surrounded by eight 1's. There is no overlap of closed rectangles.

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    int rectangle(int A[][10],int i,int j,int d1,int d2,int cnt);
    int A[][10]={
        {0,0,0,0,0,0,0,0,0,0},
        {0,0,1,1,1,1,1,0,0,0},
        {0,0,1,0,0,0,1,0,0,0},
        {0,0,1,0,0,0,1,0,0,0},
        {0,0,1,1,1,1,1,0,0,0},
        {1,1,1,0,0,0,1,1,1,1},
        {0,0,1,0,0,0,1,0,1,1},
        {1,1,1,0,0,0,1,1,1,1}
    };
    int cnt=0;
	for(int i=0;i<8;i++)
    {
        for(int j=0;j<10;j++)
        {
            cout<<A[i][j]<<" ";
        }
        cout<<endl;
    }
    for(int i=0;i<8;i++)
    {
        for(int j=0;j<10;j++)
        {
            if(A[i][j]>0)
            {
                cnt=rectangle(A,i,j,8,10,cnt);
            }
        }
    }
    cout<<"The number of closed rectangles is: "<<cnt<<endl;
}

int rectangle(int A[][10],int i,int j,int d1,int d2,int cnt)
{
    cnt++;
    int rnd=1;
    if(i<d1-1 && j<d2-1 && A[i+1][j+1]==0)
    {
		j++;
		while(j<d2 && i<d1-1 && A[i][j]==1 && A[i+1][j]==0)
		{
			A[i][j]=-1*cnt;
			j++;
		}
		A[i][j]=-1*cnt;
		rnd++;
	}
	if(i<d1-1 && j>0 && A[i+1][j-1]==0 && A[i][j]==-1*cnt)
	{
		i++;
		while(i<d1 && j>0 && A[i][j]==1 && A[i][j-1]==0)
		{
			A[i][j]=-1*cnt;
			i++;
		}
		A[i][j]=-1*cnt;
		rnd++;
	}
	if(i>0 && j>0 && A[i-1][j-1]==0 && A[i][j]==-1*cnt)
	{
		j--;
		while(j>=0 && i>0 && A[i-1][j]==0 && A[i][j]==1)
		{
			A[i][j]=-1*cnt;
			j--;
		}
		A[i][j]=-1*cnt;
		rnd++;
	}
	if(i>0 && j<d2-1 && A[i-1][j+1]==0 && A[i][j]==-1*cnt)
	{
		i--;
		while(i>=0 && j<d2-1 && A[i][j+1]==0 && A[i][j]==1)
		{
			A[i][j]=-1*cnt;
			i--;
		}
		A[i][j]=-1*cnt;
		rnd++;
	}
	if(rnd!=5 || (A[i][j]!=0 && A[i][j]!=A[i][j+1]))
	{
		cnt--;
	}
        return cnt;
}
Output:
The number of closed rectangles is: 2