在网格中对矩形进行分组 [英] Group rectangles in a grid

问题描述

我有一个随机切片的矩形网格-宽度为80单位。

I have a randomly-sliced rectangular grid - width is 80 unit.

我已经将网格的每一行的可用空间存储在这样的数组中

I already have the free spaces of each row of my grid stored in an array like this below:

[
    {pX:1,sX:15},
    {pX:30,sX:13},
    {pX:43,sX:1},
    {pX:44,sX:17}
],
[
    {pX:1,sX:15},
    {pX:16,sX:14},
    {pX:30,sX:13},
    {pX:43,sX:1},
    {pX:44,sX:17}
]

其中pX是起始点和sX代表每个矩形的宽度。

where pX is the starting point and sX represent the width of each rectangle.

一些数组项是相邻的，即 pX [i] + sX [i] = pX [i + 1] 。如何将这些数组条目组合在一起，并得到具有最大相邻宽度的结果矩形？

Some of the array entries are adjacent, i.e. pX[i]+sX[i] = pX[i+1]. How can i group these array entries together and get the resulting rectangles with the maximum adjacent width?

推荐答案

您需要收紧数组，连接相邻的线段。这段代码（ Delphi，将其视为伪代码）以所需的方式缩小数组：

You need to tighten arrays, joining adjacent segments. This code (Delp consider it as pseudocode) shrinks arrays in needed manner:

var
  pX, sX: TArray<Integer>;
  i, removed: Integer;
begin
  pX := [1, 30, 43, 44, 64, 66, 69, 72];
  sX := [15, 13, 1, 17, 2, 2, 3, 5];


  removed := 0;
  for i := 1 to High(pX) do begin
    if (pX[i - removed - 1] + sX[i - removed - 1] = pX[i]) then
    begin                ////join neighbors
      sX[i - removed - 1] := sX[i - removed - 1] + sX[i];
      Inc(removed);    ////removed++
    end
    else
    if (removed > 0) then
    begin                 ////copy to new place
      pX[i - removed] := pX[i];
      sX[i - removed] := sX[i];
    end;
  end;

  ////shorten array, remove tail
  SetLength(px, Length(pX) - removed);
  SetLength(sX, Length(sX) - removed);

  ////output result
  Memo1.Lines.Add(ArrayToString(pX));
  Memo1.Lines.Add(ArrayToString(sX));

输出

1 30 64 69 
15 31 4 8 

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