/* [ Blocks in matrix ] You are given an binary matrix filled with 0 and 1 only. We say two cells(i1, j1) and (i2, j2) are adjacent if and only if | i1 - i2 | <= 1 and | j1 - j2 | <= 1. A collection of adjacent cells forms a block. Your goal is count the number of blocks in the matrix. Input type :- T -> number of test cases n -> number of rows in matrix m -> number of column in matrix M -> matrix 1 5 5 1 1 0 0 0 0 1 0 0 1 1 0 0 1 1 0 0 1 0 0 1 0 1 0 1 Output -> number of blocks 4 1 4 4 1 0 1 1 1 1 0 1 0 0 0 0 1 1 1 1 Output-> number of blocks 2 https://www.sololearn.com/post/759002/?ref=app */ #include using namespace std; int findBlocks(vector A[], int n, int m); int main() { int T; cin>>T; while(T--) { int N,M; cin>>N>>M; vector A[N]; for(int i=0;i temp(M); A[i] = temp; for(int j=0;j>A[i][j]; } } cout< A[], vector > vis) { if(i >= 0 && j >= 0 && i < n && j < m && A[i][j] == 1 && !vis[i][j]) return true; return false; } int colnum[] = {-1, 0, 1, -1, 1, -1, 0, 1}; int rownum[] = {-1, -1, -1, 0, 0, 1, 1, 1}; void dfs(vector A[], int n, int m, int i, int j, vector >& vis){ vis[i][j] = true; for(int x=0;x<8;x++){ if(isSafe(i + rownum[x], j + colnum[x], n, m, A, vis)) dfs(A, n, m, i + rownum[x], j + colnum[x], vis); } } int findBlocks(vector A[], int n, int m) { vector > vis(n, vector(m, false)); int blocks = 0; for(int i=0;i

[Challenge] Matrix Blocks dfs