Find number of areas in a matrix -

assume have matrix :

1 1 1 0 0 0 1 1 1 0 0 1 0 0 0 0 0 1 

if 2 '1' next each other (horizontally , vertically only) , therefore belong same area. need find how many of these areas there in matrix. can see there's 2 areas of '1' in matrix. i've been trying solve hours code gets big , disgusting. there algorithms out there adopt problem ?

if don't care about:

• keeping input matrix intact

• performance , optimisations

then here's take on problem in c:

#include <stdio.h>  #define x       8 #define y       4  #define ign     1  int a[y][x] = {         { 1, 1, 1, 0, 0, 0, 0, 1 },         { 1, 1, 1, 0, 0, 1, 0, 1 },         { 0, 0, 0, 0, 0, 1, 0, 1 },         { 0, 0, 0, 0, 0, 1, 1, 1 }, };  int blank(int x, int y) {         if ((x < 0) || (x >= x) || (y < 0) || (y >= y) || (a[y][x] == 0))                 return 0;          a[y][x] = 0;          return 1 + blank(x - 1, y) + blank(x + 1, y) + blank(x, y - 1) + blank(x, y + 1); }  int main() {         int areas = 0;          int i, j = 0;          (i = 0; < x; ++i)                 (j = 0; j < y; ++j)                         if (a[j][i] == 1)                                 if (blank(i, j) > ign)                                         areas++;          printf("areas: %i\n", areas);          return 0; } 

once encounters 1, recursively expands on neighbouring 1 elements, counting them , turning them 0. if size of area larger ign, taken account.

notes:

• if need keep original matrix, have use copy input.

• if plan on using this, should turn code functions array , size parameters. global variables , algorithm implementations in main() should avoided, made exception in case because reduces complexity of code , allows algorithm more clear.

• with ign set 1, lone elements not considered area. changing ign 0 too.

• the if (a[j][i] == 1) conditional in loop not strictly necessary, serves minor optimisation avoiding unnecessary function calls, although compiler optimisations make redudant.

• you can modify listing of elements in each area.