连接组件标签-实施 [英] Connected Component Labeling - Implementation

问题描述

几天前，我曾问过类似的问题，但我还没有找到解决问题的有效方法. 我正在开发一个简单的控制台游戏，并且我有一个2D数组，如下所示:

I have asked a similar question some days ago, but I have yet to find an efficient way of solving my problem. I'm developing a simple console game, and I have a 2D array like this:

1,0,0,0,1
1,1,0,1,1
0,1,0,0,1
1,1,1,1,0
0,0,0,1,0

我正在尝试查找包含相邻1(4路连通性)的所有区域.因此，在此示例中，这两个区域如下:

I am trying to find all the areas that consist of neighboring 1's (4-way connectivity). So, in this example the 2 areas are as following:

1
1,1
  1
1,1,1,1
      1

和:

       1
     1,1
       1

我一直在努力的算法，找到了一个单元的所有邻居，并且在这种矩阵上工作得很好.但是，当我使用较大的数组(例如90 * 90)时，程序速度非常慢，有时使用的巨大数组会导致堆栈溢出.

The algorithm, that I've been working on, finds all the neighbors of the neighbors of a cell and works perfectly fine on this kind of matrices. However, when I use bigger arrays (like 90*90) the program is very slow and sometimes the huge arrays that are used cause stack overflows.

另一个问题上的一个人告诉我有关连接组件标记的问题，这是对我的问题的有效解决方案.

One guy on my other question told me about connected-component labelling as an efficient solution to my problem.

有人可以告诉我使用该算法的任何C ++代码，因为我对它如何与不相交集的数据结构一起实际工作感到困惑...

Can somebody show me any C++ code which uses this algorithm, because I'm kinda confused about how it actually works along with this disjoint-set data structure thing...

非常感谢您的帮助和时间.

Thanks a lot for your help and time.

推荐答案

我将首先为您提供代码，然后再进行一些解释:

I'll first give you the code and then explain it a bit:

// direction vectors
const int dx[] = {+1, 0, -1, 0};
const int dy[] = {0, +1, 0, -1};

// matrix dimensions
int row_count;
int col_count;

// the input matrix
int m[MAX][MAX];

// the labels, 0 means unlabeled
int label[MAX][MAX];

void dfs(int x, int y, int current_label) {
  if (x < 0 || x == row_count) return; // out of bounds
  if (y < 0 || y == col_count) return; // out of bounds
  if (label[x][y] || !m[x][y]) return; // already labeled or not marked with 1 in m

  // mark the current cell
  label[x][y] = current_label;

  // recursively mark the neighbors
  for (int direction = 0; direction < 4; ++direction)
    dfs(x + dx[direction], y + dy[direction], current_label);
}

void find_components() {
  int component = 0;
  for (int i = 0; i < row_count; ++i) 
    for (int j = 0; j < col_count; ++j) 
      if (!label[i][j] && m[i][j]) dfs(i, j, ++component);
}

这是解决此问题的常用方法.

This is a common way of solving this problem.

方向向量只是找到相邻单元(在四个方向中的每个方向)的一种好方法.

The direction vectors are just a nice way to find the neighboring cells (in each of the four directions).

dfs 函数对网格执行 depth-first-search .这只是意味着它将访问从起始单元格可到达的所有单元格.每个单元格都将标记为 current_label

The dfs function performs a depth-first-search of the grid. That simply means it will visit all the cells reachable from the starting cell. Each cell will be marked with current_label

find_components 函数遍历网格中的所有单元格，如果找到未标记的单元格(标记为1)，则开始标记组件.

The find_components function goes through all the cells of the grid and starts a component labeling if it finds an unlabeled cell (marked with 1).

这也可以使用堆栈来迭代完成. 如果将队列替换为队列，则会获得 bfs 或 breadth-first-search .

This can also be done iteratively using a stack. If you replace the stack with a queue, you obtain the bfs or breadth-first-search.

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