如何在二维矩阵中找到blob的数量?SIZE MxN
blob是连续的X
像素块。其中矩阵包含X
和O
XOOOXO
OXOXOX
XXOOXO
我想使用8邻域(见这里)。所以我希望在上面的例子中找到2个blobs。
想法很简单:标记每个连续的blob,并计算标记了多少blob。
下面是一些伪代码(你没有指定编程语言)让你开始:
numBlobs = 0;
foreach(item in matrix)
{
res = Visit(item);
if(res > 0)
{
numBlobs = numBlobs + 1;
}
}
return numBlobs;
Visit函数/方法如下:
Visit(item)
{
marked = 0;
if(item.IsX() && !IsItemMarked(neighbour))
{
MarkItemAsVisited(item);
marked = 1;
foreach(neighbour in GetNeighbours(item))
{
marked = marked + Visit(neighbour);
}
}
return marked;
}
你所要做的就是实现其他的函数/方法,但是它们非常简单。
public static void main(String[] args) {
int[][] matrix = new int[6][5];
System.out.println(matrix.length);
for (int i=0; i < matrix.length; i++) {
for (int j=0; j < matrix[i].length; j++) {
matrix[i][j] = 0;
}
}
matrix[0][3] = 1;
matrix[1][1] = 1;
matrix[1][3] = 1;
matrix[2][1] = 1;
matrix[2][2] = 1;
matrix[2][3] = 1;
matrix[4][0] = 1;
matrix[4][4] = 1;
matrix[5][2] = 1;
matrix[5][3] = 1;
matrix[5][4] = 1;
System.out.println(findBlobCount(matrix, matrix.length, matrix[0].length));
}
static int findBlobCount (int matrix[][], int rowCount, int colCount)
{
int visited[][] = new int[rowCount][colCount]; // all initialized to false
int count=0;
for (int i=0; i<rowCount; i++)
{
for (int j=0; j<colCount; j++)
{
if (matrix[i][j] == 1 && visited[i][j] == 0) // unvisited black cell
{
markVisited (i,j, matrix, visited, rowCount, colCount);
count++;
}
}
}
return count;
}
static int markVisited (int i, int j, int [][]matrix, int [][]visited, int rowCount, int colCount)
{
if (i < 0 || j < 0)
return 0;
if (i >= rowCount || j >= colCount)
return 0;
if (visited[i][j] == 1) // already visited
return 1;
if (matrix[i][j] == 0) // not a black cell
return 0;
visited[i][j] = 1;
// recursively mark all the 4 adjacent cells - right, left, up and down
return markVisited (i+1, j, matrix, visited, rowCount, colCount)
+ markVisited (i-1, j, matrix, visited, rowCount, colCount)
+ markVisited (i, j+1, matrix, visited, rowCount, colCount)
+ markVisited (i, j-1, matrix, visited, rowCount, colCount);
}