Given a 2–d matrix , which has only 1’s and 0’s in it. Find the total number of connected sets in that matrix.

Explanation:

Connected set can be defined as group of cell(s) which has 1 mentioned on it and have at least one other cell in that set with which they share the neighbor relationship. A cell with 1 in it and no surrounding neighbor having 1 in it can be considered as a set with one cell in it. Neighbors can be defined as all the cells adjacent to the given cell in 8 possible directions ( i.e N , W , E , S , NE , NW , SE , SW direction ). A cell is not a neighbor of itself.

Input format :

First line of the input contains T , number of test-cases.

Then follow T testcases. Each testcase has given format.

N [ representing the dimension of the matrix N X N ].

Followed by N lines , with N numbers on each line.

Ouput format :

For each test case print one line ,  number of connected component it has.

Sample Input :

4

4

0 0 1 0

1 0 1 0

0 1 0 0

1 1 1 1

4

1 0 0 1

0 0 0 0

0 1 1 0

1 0 0 1

5

1 0 0 1 1

0 0 1 0 0

0 0 0 0 0

1 1 1 1 1

0 0 0 0 0

8

0 0 1 0 0 1 0 0

1 0 0 0 0 0 0 1

0 0 1 0 0 1 0 1

0 1 0 0 0 1 0 0

1 0 0 0 0 0 0 0

0 0 1 1 0 1 1 0

1 0 1 1 0 1 1 0

0 0 0 0 0 0 0 0

Sample output :

1

3

3

9

Constraint :

0 < T < 6 

0 < N < 1009 

=================WORKING CODE=================

#include<stdio.h>

int count_connected_set(int A[1015][1015],int i,int j,int n)

{

    A[i][j]=2;

    if(A[i-1][j-1]==1 && i>0 && j>0 && i<=n && j<=n )     //NW

        count_connected_set(A,i-1,j-1,n);

    if(A[i-1][j]==1 && i>0 && j>0 && i<=n && j<=n)       //N

        count_connected_set(A,i-1,j,n);

    if(A[i-1][j+1]==1 && i>0 && j>0 && i<=n && j<=n)//   //NE

        count_connected_set(A,i-1,j+1,n);

    if(A[i][j-1]==1 && i>0 && j>0 && i<=n && j<=n)       //W

        count_connected_set(A,i,j-1,n);

    if(A[i][j+1]==1 && i>0 && j>0 && i<=n && j<=n)       //E

        count_connected_set(A,i,j+1,n);

    if(A[i+1][j-1]==1 && i>0 && j>0 && i<=n && j<=n)     //SW

        count_connected_set(A,i+1,j-1,n);

    if(A[i+1][j]==1 && i>0 && j>0 && i<=n && j<=n)       //S

        count_connected_set(A,i+1,j,n);

    if(A[i+1][j+1]==1 && i>0 && j>0 && i<=n && j<=n)     //SE

        count_connected_set(A,i+1,j+1,n);

return 1;

}// end of count_connected_set()

int main()

{

    int i,j,t,k,n,A[1015][1015],count;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        for(i=1;i<=n;i++)

            for(j=1;j<=n;j++)

                scanf("%d",&A[i][j]); // geting input a test case;

        count =0;

        for(i=1;i<=n;i++)

        {

            for(j=1;j<=n;j++)

                if(A[i][j]==1)

                    if(count_connected_set(A,i,j,n))

                        count++;

        }// of for

        printf("%d\n",count);

    }// end of while

return 0;

}

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