graph split （C++）

2维0，1矩阵。判断有多少个封闭的全是0的area。DFS，BFS的方法都要写。

void dfs(const vector<vector<int>> &G, int i, int j){
  if(i<0 || i >= rown || j<0 || j >= coln) return;
  if(G[i][j] == 1 || visited[i][j] == 1) return;
  visited[i][j] = 1;
  dfs(G, i - 1, j);
  dfs(G, i + 1, j);
  dfs(G, i, j+1);
  dfs(G, i, j-1);
}

int count(vector<vector<int>>G){
  if(G.empty()) return 0;
  int count;
 int rown = G.size(), coln = G[0].size;
 vector<vector<int>> visited(rown, vector<int>(coln,0));
  for(i = 0; i < rown; i++) {
    for(j = 0; j < coln; j++){
      if(visited[i][j] == 0&& G[i][j] == 0) {  
            count++;
          dfs(G,i, j, rown, coln);
        }
      }
    }

return count;

}

-------------------
void bfs(const vector<vector<int>> &G, int i, int j){
  queqe<pair<int,int>> q;
  q.push(make_pair(i,j));
  while(!q.empty()){
     i = queue.front().first; j = queue.front().second;
     queue.pop();
     visited[i][j] = 1;
     if(i > 0 && visited[i-1][j] == 0 && G[i-1][j] == 0) queue.push(make_pair(i-1,j));
     if(j> 0 && visited[i][j-1] == 0 && G[i][j-1] == 0) queue.push(make_pair(i,j-1));
     if(i < rown - 1 && visited[i+1][j] == 0 && G[i+1][j] == 0) queue.push(make_pair(i+1,j));
     if(j < coln - 1 && visited[i][j+1] == 0&& G[i][j+1] == 0) queue.push(make_pair(i,j+1));
  }
 

}


int count(vector<vector<int>>G){
  if(G.empty()) return 0;
  int count;
 int rown = G.size(), coln = G[0].size;
 vector<vector<int>> visited(rown, vector<int>(coln,0));
  for(i = 0; i < rown; i++) {
    for(j = 0; j < coln; j++){
      if(visited[i][j] == 0&& G[i][j] == 0) {  
            count++;
          bfs(G,i, j, rown, coln);
        }
      }
    }

return count;

}