查找岛屿数量的 C 程序 |设置 1（使用 DFS）

给定一个布尔二维矩阵，找出岛屿的数量。一组连接的 1 形成一个岛。例如，下面的矩阵包含 5 个岛屿

例子：

Input : mat[][] = {{1, 1, 0, 0, 0},
                   {0, 1, 0, 0, 1},
                   {1, 0, 0, 1, 1},
                   {0, 0, 0, 0, 0},
                   {1, 0, 1, 0, 1} 
Output : 5

这是标准问题的变体：“计算无向图中的连通分量的数量”。
在我们解决问题之前，让我们了解什么是连接组件。无向图的连通分量是一个子图，其中每两个顶点通过一条路径相互连接，并且不与子图之外的其他顶点相连。
例如，下图具有三个连通分量。

C/C++

// Program to count islands in boolean 2D matrix
#include 
#include 
#include 
  
#define ROW 5
#define COL 5
  
// A function to check if a given cell (row, col) can be included in DFS
int isSafe(int M[][COL], int row, int col, bool visited[][COL])
{
    // row number is in range, column number is in range and value is 1 
    // and not yet visited
    return (row >= 0) && (row < ROW) &&     
           (col >= 0) && (col < COL) &&      
           (M[row][col] && !visited[row][col]); 
}
  
// A utility function to do DFS for a 2D boolean matrix. It only considers
// the 8 neighbours as adjacent vertices
void DFS(int M[][COL], int row, int col, bool visited[][COL])
{
    // These arrays are used to get row and column numbers of 8 neighbours 
    // of a given cell
    static int rowNbr[] = {-1, -1, -1,  0, 0,  1, 1, 1};
    static int colNbr[] = {-1,  0,  1, -1, 1, -1, 0, 1};
  
    // Mark this cell as visited
    visited[row][col] = true;
  
    // Recur for all connected neighbours
    for (int k = 0; k < 8; ++k)
        if (isSafe(M, row + rowNbr[k], col + colNbr[k], visited) )
            DFS(M, row + rowNbr[k], col + colNbr[k], visited);
}
  
// The main function that returns count of islands in a given boolean
// 2D matrix
int countIslands(int M[][COL])
{
    // Make a bool array to mark visited cells.
    // Initially all cells are unvisited
    bool visited[ROW][COL];
    memset(visited, 0, sizeof(visited));
  
    // Initialize count as 0 and traverse through the all cells of
    // given matrix
    int count = 0;
    for (int i = 0; i < ROW; ++i)
        for (int j = 0; j < COL; ++j)
            if (M[i][j] && !visited[i][j]) // If a cell with value 1 is not
            {                         // visited yet, then new island found
                DFS(M, i, j, visited);     // Visit all cells in this island.
                ++count;                   // and increment island count
            }
  
    return count;
}
  
// Driver program to test above function
int main()
{
    int M[][COL]= {  {1, 1, 0, 0, 0},
        {0, 1, 0, 0, 1},
        {1, 0, 0, 1, 1},
        {0, 0, 0, 0, 0},
        {1, 0, 1, 0, 1}
    };
  
    printf("Number of islands is: %d\n", countIslands(M));
  
    return 0;
}

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