Thread: Find the number of islands in a given matrix and area of every island

Dear,

I am counting the number of islands in a given matrix and try to find the area of every island.
http://en.wikipedia.org/wiki/Connect…raph_theory%29
Now finally remove the errors and my result is
1 islands area is 1
2 islands area is 1
3 islands area is 1
4 islands area is 1
5 islands area is 1
Number of islands is: 5
But I need a result according to my matrix is
1 islands area is 4
2 islands area is 3
3 islands area is 1
4 islands area is 1
5 islands area is 1
Number of islands is: 5
1-I am thinking that there is some problem for calling the area.
what do you think?
Please reply me

Code:

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include<conio.h>
#define ROW 5
#define COL 5
int markIsland(int arr[][COL], int i, int j, int m, int n,int area);
 int countIslands(int arr[][COL], int m, int n);
 int markIsland(int arr[][COL], int i, int j, int m, int n,int area)
 {
 // area=1;
{
arr[i][j] = -1;
 if(i-1 >= 0)
{
 if(j-1 >= 0 && arr[i-1][j-1] == 1)
markIsland(arr, i-1, j-1, m, n,area);
 if(arr[i-1][j] == 1)
markIsland(arr, i-1, j, m, n,area);
 if(j+1 < n && arr[i-1][j+1] == 1)
markIsland(arr, i-1, j+1, m, n,area);
 }
 if(i+1 < m)
{
 if(j-1 >= 0 && arr[i+1][j-1] == 1)
markIsland(arr, i+1, j-1, m, n,area);
 if(arr[i+1][j] == 1)
markIsland(arr, i+1, j, m, n,area);
 if(j+1 < n && arr[i+1][j+1] == 1)
markIsland(arr, i+1, j+1, m, n,area);
 }
 if(j-1 >= 0 && arr[i][j-1] == 1)
markIsland(arr, i, j-1, m, n,area);
 if(j+1 < n && arr[i][j+1] == 1)
markIsland(arr, i, j+1, m, n,area);
 area++;
 }
return area;
}
 int countIslands(int arr[][COL], int m, int n)
{
int count = 0;
 for(int i=0; i&lt;m; i++)
{
for(int j=0; j&lt;n; j++)
 if(arr[i][j] == -1)
arr[i][j] = 1;
 count++;
 }
 return count;
}
 main()
 {
int area=1,count=1;
int v, i,j;
int m = ROW;
int n = COL;
int arr[][COL]= { {1, 1, 0, 0, 0},
{0, 1, 0, 0, 1},
{1, 0, 0, 1, 1},
{0, 0, 0, 0, 0},
{1, 0, 1, 0, 1}
};
for(v=0;v&lt; countIslands(arr,m,n);v++)
{
printf(&quot; %d islands area is %d\n&quot;, count++,area);
// printf(&quot; %d islands area is %d\n&quot;, countIslands(arr,m,n),markIsland(arr,i,j,m,n,area) );
}
printf(&quot;Number of islands is: %d\n&quot;, countIslands(arr,m,n));
 getch();
}

I see that cross post Salem, and I raise you one:
Find The Number Of Islands In A Given Matrix And Area Of Every Island - C And C++ | Dream.In.Code

And the link still doesn't work.

Connected component (graph theory) - Wikipedia, the free encyclopedia

what about the code?

Originally Posted by Adak

I see that cross post Salem, and I raise you one:
Find The Number Of Islands In A Given Matrix And Area Of Every Island - C And C++ | Dream.In.Code

And the link still doesn't work.

Dear,
Thanks for ur reply,
Now there is few errors in my code,why they give me TOO FEW ARGYMENTS error?
FOr example:
int count_connected_set(int (*)[5], int, int, int, int, int, int)':
invalid conversion from `int' to `int (*)[5]'
too few arguments to function `int countIslands(int (*)[5], int, int)'
int main()':

Code:

 #include<stdio.h>
#include<conio.h>
#define ROW 5
#define COL 5

int count_connected_set(int A[][COL], int i, int j,int m,int n,int area,int count);
int countIslands(int A[][COL], int m, int n);
int area=0;
int i=1;
int j=1;
int count=0;
int count_connected_set(int A[][COL],int i,int j,int m,int n,int area,int count)
{
     A[i][j] = -1;
  
     {
    if(A[i-1][j-1]==1 && i>0 && j>0 && i<=m && j<=n )     //NW
       
        count_connected_set(A,i-1,j-1,m,n,area,count);
        area += count_connected_set(A, i-1, j-1, m, n);
       
    if(A[i-1][j]==1 && i>0 && j>0 && i<=m && j<=n)       //N
       
        count_connected_set(A,i-1,j,m,n,area,count);
        area += count_connected_set(A, i-1, j, m, n);
      
    if(A[i-1][j+1]==1 && i>0 && j>0 && i<=m && j<=n)//   //NE
        
        count_connected_set(A,i-1,j+1,m,n,area,count);
        area += count_connected_set(A, i-1, j+1, m, n);
        
    if(A[i][j-1]==1 && i>0 && j>0 && i<=m && j<=n)       //W
        
        count_connected_set(A,i,j-1,m,n,area,count);
        area += count_connected_set(A, i, j-1, m, n);
        
    if(A[i][j+1]==1 && i>0 && j>0 && i<=m && j<=n)       //E
        
        count_connected_set(A,i,j+1,m,n,area,count);
        area += count_connected_set(A, i, j+1, m, n);
        
    if(A[i+1][j-1]==1 && i>0 && j>0 && i<=m && j<=n)     //SW
        
        count_connected_set(A,i+1,j-1,m,n,area,count);
        area += count_connected_set(A, i+1, j-1, m, n);
       
    if(A[i+1][j]==1 && i>0 && j>0 && i<=m && j<=n)       //S
       
        count_connected_set(A,i+1,j,m,n,area,count);
        area += count_connected_set(A, i+1, j, m, n);
       
    if(A[i+1][j+1]==1 && i>0 && j>0 && i<=m && j<=n)     //SE
       
        count_connected_set(A,i+1,j+1,m,n,area,count);
        area += count_connected_set(A, i+1, j+1, m, n);
      
       }

       return area;
 }
// end of count_connected_set()

int countIslands(int A[][COL], int m, int n)
       {
       int count = 0;
     
       for(int i=0; i<m; i++)
               {     
               for(int j=0; j<n; j++)
                       {
                        if(A[i][j] == -1)
                        A[i][j] = 1;
              
                 count_connected_set(A, i, j, m, n,area);
                 count++;
                 
                          }
              }
       return count;
       }

main()

      {
     
      int  i,j;
      int m = ROW;
      int n = COL;
      int A[][COL]= { {1, 1, 0, 0, 0},
                        {0, 1, 0, 0, 1},
                        {1, 0, 0, 1, 1},
                        {0, 0, 0, 0, 0},
                        {1, 0, 1, 0, 1}
                      };
                      
                      
              
          
              printf(" %d islands area is %d\n", count_connected_set(count),count_connected_set (area));
                      
                       printf("Number of islands is: %d\n",countIslands(count));
      getch();
      }

Originally Posted by Salem

What about the code!?

What about all the replies you've already got on other forums?
How about reading all of those, making some changes to the code and then posting back (presumably to all the forums you've cross-posted to already) what your new code is.

Or perhaps you're just hoping that if you post the same old rubbish to enough forums, one will eventually post a completely correct answer without you having to do any more work. If this is your idea, please leave.

@salem bhai.....No i am try to solve the problem and posting to other forum only for getting more ideas for optimize the code.
Therefore I change the code due to errors.
OK,next time I will post only this forum .

Thanks
but still `int count_connected_set(int (*)[5], int, int, int, int, int, int)'

But I want to call only area,how is it possible program give me area value ?



At

Originally Posted by rogster001

if you declare a function having 7 arguments, then you must call it with 7, unless it can be overloaded

is it correct way ?
printf(" %d islands area is %d\n", count_connected_set(A, i, j, m, n,area,count),count_connected_set(A, i, j, m, n,area,count));

printf("Number of islands is: %d\n",countIslands( A, m, n,count));

Originally Posted by tahir123

but still `int count_connected_set(int (*)[5], int, int, int, int, int, int)'

But I want to call only area,how is it possible program give me area value ?

You have written the program (haven't you?) so you should know what alle these int arguments should be and how to call the function.

Bye, Andreas

Originally Posted by tahir123

is it correct way ?

Does it compile?

If you get errors it can't be correct.

Bye, Andreas

@Andreas.........no still i got the error therefore i change the arguments.

I think we all knew this, but the OP did not write the code themselves. It looks like they took it from here: CODES: matrix.

@tahir123:
Do your own work, and you might stand a chance of actually understanding what is going wrong, and you just might be able to fix it. We also might be willing to help, if you actually tried anything other than plagiarism.