查找岛屿数量的PHP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。
PHP
= 0) && ($row < $ROW) &&
($col >= 0) && ($col < $COL) &&
($M[$row][$col] &&
!isset($visited[$row][$col]));
}
// A utility function to do DFS
// for a 2D boolean matrix. It
// only considers the 8 neighbours
// as adjacent vertices
function DFS(&$M, $row, $col,
&$visited)
{
// These arrays are used to
// get row and column numbers
// of 8 neighbours of a given cell
$rowNbr = array(-1, -1, -1, 0,
0, 1, 1, 1);
$colNbr = array(-1, 0, 1, -1,
1, -1, 0, 1);
// Mark this cell as visited
$visited[$row][$col] = true;
// Recur for all
// connected neighbours
for ($k = 0; $k < 8; ++$k)
if (isSafe($M, $row + $rowNbr[$k],
$col + $colNbr[$k], $visited))
DFS($M, $row + $rowNbr[$k],
$col + $colNbr[$k], $visited);
}
// The main function that returns
// count of islands in a given
// boolean 2D matrix
function countIslands(&$M)
{
global $ROW, $COL;
// Make a bool array to
// mark visited cells.
// Initially all cells
// are unvisited
$visited = array(array());
// Initialize count as 0 and
// traverse through the all
// cells of given matrix
$count = 0;
for ($i = 0; $i < $ROW; ++$i)
for ($j = 0; $j < $COL; ++$j)
if ($M[$i][$j] &&
!isset($visited[$i][$j])) // If a cell with value 1
{ // is not visited yet,
DFS($M, $i, $j, $visited); // then new island found
++$count; // Visit all cells in this
} // island and increment
// island count.
return $count;
}
// Driver Code
$M = array(array(1, 1, 0, 0, 0),
array(0, 1, 0, 0, 1),
array(1, 0, 0, 1, 1),
array(0, 0, 0, 0, 0),
array(1, 0, 1, 0, 1));
echo "Number of islands is: ",
countIslands($M);
// This code is contributed
// by ChitraNayal
?>
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