Php 所以，最后，你会得到： 这就产生了解决方案：这些数字序列（绿色区域）是两个可能的矩形3x5的自由座位的上边缘 该算法可以很容易地增强，例如，获得所有具有最大面积的矩形。或者，它可以用来找到N个座位的任何连续区域（不仅仅是矩形区域）——只是在第二遍中，寻找

所以，最后，你会得到：

这就产生了解决方案：这些数字序列（绿色区域）是两个可能的矩形3x5的自由座位的上边缘

---

该算法可以很容易地增强，例如，获得所有具有最大面积的矩形。或者，它可以用来找到N个座位的任何连续区域（不仅仅是矩形区域）——只是在第二遍中，寻找至少等于N的连续数字序列。

我来这里是为了寻找Python解决方案。这是我的，它返回所有位置：

import numpy

s = '''0 0 0 0 1 0 0 0 1 0 0
0 0 0 0 0 0 1 0 1 0 1
0 1 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0'''

area = 15
nrows = 6
ncols = 11
skip = 1

a = numpy.fromstring(s, dtype=int, sep=' ').reshape(nrows, ncols)
w = numpy.zeros(dtype=int, shape=a.shape)
h = numpy.zeros(dtype=int, shape=a.shape)
for r in range(nrows):
    for c in range(ncols):
        if a[r][c] == skip:
            continue
        if r == 0:
            h[r][c] = 1
        else:
            h[r][c] = h[r-1][c]+1
        if c == 0:
            w[r][c] = 1
        else:
            w[r][c] = w[r][c-1]+1
        minw = w[r][c]
        for dh in range(h[r][c]):
            minw = min(minw, w[r-dh][c])
            if (dh+1)*minw == area:
                print(
                    'area', area, 'row', r, 'col', c,
                    'height', dh+1, 'width', minw)

输出：

area 15 row 4 col 8 height 3 width 5
area 15 row 5 col 10 height 3 width 5

---

为此干杯。。。伟大的信息调查，因为你可以理解的复杂性（和非常繁忙），它可能需要一些时间来验证结果！同时我也投了赞成票，我不认为在DB这样做是个好主意。场地小到可以更快、更高效地将其加载到内存中并在那里执行操作，而不是执行磁盘操作。所有这些都过于复杂。在O（N）中有一个非常简单的解决方案，见我的帖子。顺便说一句，只是一个“草稿”，但是你得到了这样的想法。你已经写过类似这样的东西了-积木可能是弯曲的或梯形的，前面的座位比后面的座位要少，这对于概念验证来说是很好的，但是如果你处理了100或1000次预订请求，这是不可行的：）我不确定我是否知道你在尝试做什么以及情况如何。。。但是，应用一排可用席位的掩码（如IP子网掩码）怎么样？e、 g.

0001111000

，所有的1都可用。布赖恩，你想找到所有的矩形块3x5的免费座位吗？这是正确的吗？我发布了一个答案-这是你想要的吗？你可以修改得很好，因为简单的解决方案（尝试每个边界框）至少是O（nm），其中m是块的大小。不过，你怎么才能找到一排7人的座位呢？@Nick，非常简单——在第二遍中，只需寻找至少7人的数字序列即可。例如，您可以添加“不超过两行”这样的条件，因此在这种情况下，每个大于2的数字将仅计为2。项目已被合并。。。但这将是一个优雅的解决方案。

SELECT    a.event_id, a.performance, a.block,
          a.row, a.seat AS start_seat,
          a.seat + (4 - 1) AS end_seat,
          4 AS requested_seats,
          a.id AS start_allocation_id
FROM      seats a
          LEFT JOIN seats b ON
              a.event_id = b.event_id AND
              a.performance = b.performance AND
              a.block = b.block AND
              a.row = b.row AND
              a.seat < b.seat AND
              b.seat < a.seat + 4 AND
              b.status = 1
WHERE     a.status = 1 AND
          a.event_id = 1
GROUP BY  a.seat
HAVING COUNT(b.seat) + 1 = 4
ORDER BY performance

    $nr_tickets = 15; // or whatever


// build an array of different configurations:
// since you always want people as close to eachother as possible this is a suggestion:

$configurations = array();
for($columns=1; $columns<=$nr_tickets; $columns++)
{
    $BLOCK = array();
    $remainder = $nr_tickets % $columns;
    // $nr_tickets - $remainder = greatest number divisible exactly by $i (columns) which is the number of rows you want.
    $rows = (($nr_ticket-$odd) / $i);
    //$configurations[$columns] = $rows // make a rectangle configuration or $columns x $rows with $remainder tickets left.
    $BLOCK[] = array_fill(0, $columns, array_fill(0, $rows, X); // multidimensional array
    for($a=0; $a<$odd; $a++)
    {

        // each of the remainder seats need to be 'stuck on to the block/rectangle of seats you already have, this can be done in
        // ($rows + $columns * 2) - $a ways for each of the consecutive non-assigned tickets
        /*
            lets say you have a block of 2x7 (s) with 1 (N) possible ticket left 
                   N  N  N  N  N  N  N  
            N  s  s  s  s  s  s  s  N 
            N  s  s  s  s  s  s  s  N   
               N  N  N  N  N  N  N 
        */
        // so basically we can add a ticket to each of the rows and for each of those configurations we need to add $a more
        // this may be simpler with a recursive function call that adds $a more for each 1 ticket you add here to a row or a column. 
    }


}



// now you can go select all different permutations of settings from the database and select one you like :) 

CREATE TABLE Seat {
   SeatID int,
   Status int,
   ...
   NorthID int,
   NorthWestID int,
   WestID int,
   ...
   NorthEastID int
}

SELECT * FROM Seat x
INNER JOIN Seat n ON x.NorthID = n.SeatID
INNER JOIN Seat nw ON x.NorthWestID = n.SeatID
...

SELECT row, col,
       (select sum(taken) from seat as s2
               where s2.row >= s1.row and s2.row < s1.row + 2
                 and s2.col >= s1.col and s2.col < s1.col + 2) as count
       FROM seat as s1

import numpy

s = '''0 0 0 0 1 0 0 0 1 0 0
0 0 0 0 0 0 1 0 1 0 1
0 1 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0'''

area = 15
nrows = 6
ncols = 11
skip = 1

a = numpy.fromstring(s, dtype=int, sep=' ').reshape(nrows, ncols)
w = numpy.zeros(dtype=int, shape=a.shape)
h = numpy.zeros(dtype=int, shape=a.shape)
for r in range(nrows):
    for c in range(ncols):
        if a[r][c] == skip:
            continue
        if r == 0:
            h[r][c] = 1
        else:
            h[r][c] = h[r-1][c]+1
        if c == 0:
            w[r][c] = 1
        else:
            w[r][c] = w[r][c-1]+1
        minw = w[r][c]
        for dh in range(h[r][c]):
            minw = min(minw, w[r-dh][c])
            if (dh+1)*minw == area:
                print(
                    'area', area, 'row', r, 'col', c,
                    'height', dh+1, 'width', minw)

area 15 row 4 col 8 height 3 width 5
area 15 row 5 col 10 height 3 width 5