【发布时间】:2016-10-24 03:37:54
【问题描述】:
我似乎在绘制正确的十六进制网格时遇到了一些麻烦:
如您所见,六边形只是略微未对齐,但我相信我的数学是正确的(其中一些已通过 http://www.redblobgames.com/grids/hexagons/ 验证)。
我的绘制方法是从左上角的六边形(第一行的第一个瓦片)开始画那一行瓦片。然后对于下一行,有一个负的 X 偏移和正的 Y 偏移等,直到它到达中间行,X 偏移增加直到 0:
private function drawHexGrid(inputGraphics:Graphics, inputPos:Point, inputGrid:HexGrid, inputTileSize:int):void {
var rootThree:Number = Math.sqrt(3); // seems like this will be used a lot
// total number of rows and also size of largest row (in tiles)
var totalRows:int = (2 * inputGrid.size) - 1;
// the other useful dimension of a hex tile
var triSize:Number = rootThree * 0.5 * inputTileSize;
var topLeft:Point = new Point(-(inputGrid.size - 1) * triSize, -(1.5 * (inputGrid.size - 1) * inputTileSize));
topLeft.x += inputPos.x;
topLeft.y += inputPos.y;
var currentPos:Point = topLeft.clone();
// step size between each tile and row
var xStep:Number = rootThree * inputTileSize;
var yStep:Number = triSize * rootThree;
var offsetDirection:int = -1;
var rowLimit:int = inputGrid.size;
var offsetAmount:int = 0; // offsetAmount goes 1 to n, then back to 0 again, used for calculating thw row offsets
var mazeTiles:Vector.<Tile> = inputGrid.getTiles();
var tileCounter:int = 0; // index to cycle through mazeTiles
for(var rowCount:int = 0; rowCount < totalRows; rowCount++){
currentPos.x = topLeft.x + (offsetAmount * rootThree / -2 * inputTileSize);
for(var counter:int = 0; counter < rowLimit; counter++){
drawHexTile(inputGraphics, currentPos.x, currentPos.y, inputTileSize, mazeTiles[tileCounter++]);
currentPos.x += xStep;
}
currentPos.y += yStep;
if(rowCount == (inputGrid.size - 1)){
offsetDirection *= -1;
}
rowLimit += offsetDirection * -1;
offsetAmount -= offsetDirection;
} // end of for loop
} // end of drawHexGrid()
每个六边形的实际绘制都在这个循环中:
private function drawHexTile(inputGraphics:Graphics, inputX:int, inputY:int, inputSize:int, inputTile:Tile):void {
inputGraphics.lineStyle(0.1, 0, 1);
var convertToRadians:Number = Math.PI / 180;
// easier to draw by wall, since need to look up each wall and can set a starting degree without having to worry about the 'end degree'
// (since the end may not be 360 degrees if the starting degree is in the negatives or a high degree)
var degrees:int = -150; // starting wall is the top left wall of the hexagon tile
for(var counter:int = 0; counter < 6; counter++){
if(inputTile.walls[counter] == true){
inputGraphics.moveTo(inputX + (Math.cos(degrees * convertToRadians) * inputSize),
inputY + (Math.sin(degrees * convertToRadians) * inputSize));
inputGraphics.lineTo(inputX + (Math.cos((degrees + 60) * convertToRadians) * inputSize),
inputY + (Math.sin((degrees + 60) * convertToRadians) * inputSize));
}
degrees += 60;
}
} // end of drawHexTile() method
乍一看,我认为问题在于浮点数学准确性,但我不确定如何修复/改进此问题。
有什么想法吗?
我知道我的绘图方法最终会产生许多重叠的线条,目前这很好。如果有帮助,整个代码适用于六边形迷宫生成器,它工作正常,因此可以忽略,因为它不是问题的一部分。
如果有帮助,我如何存储六边形图块只是在一个长数组中,索引如下:
where n=4
0 1 2 3
4 5 6 7 8
9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27
28 29 30 31 32
33 34 35 36
【问题讨论】:
-
虽然不适合您的最终产品,但请尝试将所有 int 值更改为 Num 值。走着瞧吧。还要关注处理每个新图块的 X 位置的代码。
-
……呃……这行得通。我不知道为什么。 =|我刚刚对 Numbers 的整数进行了全局替换,它起作用了。嗯,将返回并手动替换以查看确切的 int 变量。
-
哦!我想到了。在 drawHexTile() 中,我将 inputX 和 inputY 设置为 int 类型。但实际上,计算的位置是浮动值,因此它们被四舍五入到整数,这隐含地导致了错位。简单的错误(通过反射),毕竟修复中没有涉及任何魔法。如果您将您的评论重新发布为答案,我会将其标记为已接受。
标签: actionscript-3 drawing hexagonal-tiles