2 回答

TA貢獻1757條經驗 獲得超7個贊
所以這真的像一個充洪問題。您有一個固定的 2D 格網作為數組,并且您希望根據 X/Y 坐標填充特定區域。我簡化了下面的實現,以演示如何完成此操作。
const GRID_WIDTH = 10;
const GRID_HEIGHT = 10;
// Coordinates are described as [[y1, x1], [y2, x2]]
$coordinates = [[1,3], [3,3]];
// Create the grid
$grid = array_fill(0, GRID_WIDTH, array_fill(0, GRID_HEIGHT, '-'));
// Plot the coordinates on the grid
for ($y = 0; $y < GRID_WIDTH; $y++) { // plot Y
for ($x = 0; $x < GRID_HEIGHT; $x++) { // plot X
if ($y >= $coordinates[0][0] && $y <= $coordinates[0][1]) {
if ($x >= $coordinates[1][0] && $x <= $coordinates[1][1]) {
$grid[$y][$x] = 'X';
}
}
}
}
// Print the grid
foreach ($grid as $x => $y) {
foreach ($y as $plot) {
echo " $plot ";
}
echo "\n";
}
輸出應如下所示:
- - - - - - - - - -
- - - X - - - - - -
- - - X - - - - - -
- - - X - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
在你的第一個例子中,它看起來像這樣:[0,1][1,2]
- X X - - - - - - -
- X X - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
- - - - - - - - - -
這里的關鍵是確保當我們掃描網格(從X到Y)時,X1,Y1,X2和Y2的坐標在光標的范圍內。這允許我們一次繪制一個單元格,只要我們在這些邊界內(包括)。這是通過檢查 和 來完成的。由于計算坐標的寬度是不夠的(這就是上面的實現正在做的事情)。因為它不標識高度邊界。if ($y >= $coordinates[0][0] && $y <= $coordinates[0][1])if ($x >= $coordinates[1][0] && $x <= $coordinates[1][1])

TA貢獻1860條經驗 獲得超8個贊
您可以使用已有的代碼(盡管不需要計算)迭代要繪制的矩形:$width
$rects = array(array(array(1, 3), array(3, 3)),
array(array(0, 1), array(1, 2))
);
$grid = array_fill(0, 10, array_fill(0, 10, '0'));
foreach ($rects as $r) {
list($upperLeftNode, $lowerRightNode) = $r;
for($x = $upperLeftNode[0]; $x <= $lowerRightNode[0]; $x++) {
for($y = $upperLeftNode[1]; $y <= $lowerRightNode[1]; $y++) {
$grid[$x][$y] = 'x';
}
}
}
foreach ($grid as $row) {
echo implode(' ', $row) . "\n";
}
輸出:
0 x x 0 0 0 0 0 0 0
0 x x x 0 0 0 0 0 0
0 0 0 x 0 0 0 0 0 0
0 0 0 x 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
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