芬兰电影院的座位

52

您会得到一个布尔值矩阵的电影院地图：0代表一个免费座位，1代表已占用。每个进入的芬兰人都从最近的被占领者那里选择最远的座位（欧几里得距离），如果有多个座位的话，则按行顺序排列。输出一个矩阵，显示最终将占用的订单；也就是说，将0替换为2、3、4等

// in
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
0 0 1 1 0
// out
 2  8  3  9  1
10  5 11  6 12
 4 13 14 15  7
16 17  1  1 18

// in
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 1 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 1 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 1 0 0 0
// out
  5  43  17  44  45  46  18  47   8  48  49   6  50  19  51   2
 52  24  53  54   1  55  56  25  57  26  58  59  27  60  28  61
 20  62  63  29  64  65   1  66  30  67  68  21  69   9  70  71
 72  73   1  74  31  75  76  77  78   1  79  80  32  81  82  11
 12  83  84   1  85  86  87  13  88  89  90  14  91  92  33  93
 94  34  95  96  97  15  98  99  35 100  36 101 102   1 103  22
104 105  37 106  38 107  39 108 109  16 110  40 111 112  41 113
  4 114 115   7 116  23 117   3 118 119  42 120   1 121 122  10

// in
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 1 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
// out
  2  38 39  26  40   6 41  42  12  43  44   7  45  46  27  47   3
 48  49 15  50  28  51 52  29  53  30  54  55  56  16  57  31  58
 32  59 60  33  61  62 17  63  64  65  18  66  67  68  34  69  35
 70  10 71  72  13  73 74  75   1  76  77  78  11  79  80  14  81
 82  83 36  84  85  86 21  87  88  89  22  90  91  37  92  93  94
 19  95 96  97  23  98 99 100  24 101 102 103  25 104 105 106  20
107 108  4 109 110 111  8 112 113 114   9 115 116 117   5 118 119

在您所用语言的既定代码高尔夫规范中，I / O格式很灵活。您可以假设输入是正确的，大小至少为3x3，并且不完全由相同的布尔值组成。编写函数或完整程序。每种语言中最短的解决方案被认为是赢家；没有答案将被接受。禁止出现标准漏洞。

code-golf  matrix 
— ngn
source
6
@Mego作为反社会人士，我可以确定我宁愿在一个人的侧面坐两个座位，也可以在一个人后面坐两个座位，而不是在对角后面和侧面坐一个座位。
— 帕维尔
17
@Mego个人空间的计算与欧氏距离:)
— Angs
2
@Pavel Asocial，不是反社会的吗？
— Chromatix
2
@Chromatix不。我希望社会燃烧。：P
— Pavel
12
@Pavel infernosocial :)
— NGN

Answers:

11

MATL，37字节

!t~z:Q"@yX:gG&n:!J*w:+X:&-|w/X<&X>(]!

在线尝试！或验证所有测试用例。您可能还希望看到电影院充满了 ASCII艺术。

说明

!t        % Implicit input: M×N matrix of zeros and ones. Transpose and duplicate.
          % The transpose is needed because MATL uses column-major (not row-major)
          % order. It will be undone at the end
~z        % Number of zeros, say Z
:Q        % Range, add 1 element-wise: gives the array [2, 3, ..., Z+1]. These are
          % the new values that will be written into the matrix
"         % For each k in that array
  @       %   Push k. Will be written in a position to be determined
  y       %   Duplicate from below: pushes a copy of the current matrix, that has
          %   values up to k-1 already written in
  X:      %   Linearize into an (R*C)×1 vector, in column-major order
  g       %   Convert to logical: this replaces non-zero values by 1 
  G&n     %   Push input size as two separate numbers: M, N
  :!      %   Range, transpose: gives the column vector [1; 2; ...; N]
  J*      %   Multiply by imaginary unit, 1j, element-wise
  w:      %   Swap, range: gives the row vector [1, 2, ..., M]
  +       %   Add, with broadcast. Gives an N×M complex matrix defining a grid of
          %   coordinates: [1+1j, ..., M+1j; 2+1j, ... 2+1j; ...; N+1j, ..., N+Mj]
  X:      %   Linearize into an (M*N)×1 vector, in column-major order
  &-|     %   (M*N)×(M*N) matrix of absolute differences. This gives all distances
          %   between seats. Rows of this matrix represent currently used seats,
          %   and columns correspond to potential new positions
  w/      %   Swap, divide with broadcast. This divides the rows representing
          %   occupied seats by 1, and those with unocuppied seats by 0. So the
          %   latter rows are set to infinity, which effectively removes them for
          %   the subsequent minimization
  X<      %   Mimimum of each column: this gives the minimum distance to currently
          %   occupied seats for each potential new seat
  &X>     %   Argument maximum: gives the index of the first maximizing value
  (       %   Write value k at that position, using linear indexing
]         % End
!         % Transpose. Implicit display
— 路易斯·门多
source
11

的JavaScript（ES6），156个 137字节

@ l4m2节省了18个字节

相当多map()...

f=(a,n=1)=>a.map(B=(r,y)=>r.map((_,x)=>a.map(b=q=>q.map(v=>b=b<(d=X*X--+Y*Y)|!v?b:d,X=x)&Y--,Y=y)|v|b<=B||(R=r,C=x,B=b)))|B?f(a,R[C]=++n):a

在线尝试！

已评论

f = (a, n = 1) =>               // a = input array; n = seat counter
  a.map(B =                     // initialize B to a non-numeric value
    (r, y) =>                   // for each row r at position y in a[]:
    r.map((_, x) =>             //   for each target seat at position x in r[]:
      a.map(b =                 //     initialize b to a non-numeric value
        q =>                    //     for each row q in a[]:
        q.map(v =>              //       for each reference seat v in q[]:
          b = b < (             //         if b is less than d, defined as
            d = X * X-- + Y * Y //           the square of the Euclidean distance
          ) | !v ?              //           or the reference seat is empty
            b                   //             let b unchanged
          :                     //           else:
            d,                  //             update b to d
          X = x                 //         start with X = x
        ) & Y--,                //       end of q.map(); decrement Y
        Y = y                   //       start with Y = y
      ) |                       //     end of inner a.map()
      b <= B ||                 //     unless b is less than or equal to B,
      (R = r, C = x, B = b)     //     update B to b and save this position in (R, C)
    )                           //   end of r.map()
  ) | B ?                       // end of outer a.map(); if B was updated:
    f(a, R[C] = ++n)            //   update the best target seat and do a recursive call
  :                             // else:
    a                           //   stop recursion and return a[]
— Arnauld
source
— l4m2
b=b<(d=X*X--+Y*Y)|!v?b:d
— l4m2
v|b<=B v|如果是v则是不必要的b=0
— l4m2 '18
3

Haskell中，216 213 185 184个字节

import Data.Array
m a=[snd$maximum a|a/=[]]
f k=k//m[(r,((a,b),maximum(elems k)+1::Int))|s<-[assocs k],((a,b),0)<-s,r<-[minimum[(x-a)^2+(y-b)^2|((x,y),i)<-s,i>0]]]
(until=<<((==)=<<))f

将输入作为数组。输入和输出的顺序相反。归功于莱科尼的定点魔术。

在线尝试！

— 昂斯
source
1
180个字节与until((==)=<<f)f
— OVS
3

Python 2中，200个 187字节

a=input()
z=len(a[0]);P=[divmod(i,z)for i in range(len(a)*z)];i=2
while 0in sum(a,[]):t,y,x=max((min((u-U)**2+(v-V)**2for V,U in P if a[V][U]),-v,-u)for v,u in P);a[-y][-x]=i;i+=1
print a

在线尝试！

通过删除不需要的单元格为0 ，从Not that Charles到-13字节的提示。

— 查斯·布朗
source
我有几乎相同的解决方案，但是Python3，一个函数和194个字节
— 不是那位Charles
除了节省我的工作量之外，主要的节省是我将其添加,v,u到了generator的内部inside max，并且您不需要这样做，if a[v][u]<1因为这些将是0最大的，因此不是max。所以我的台词基本上是*_,y,x=max((min(...),-v,-u,v,u)for v,u in P)
— 不是查尔斯（Charles）
但是，惊人的相似代码。哇。
— 并不是说查尔斯
老实说，我不确定添加的*,v,u字符与--您拥有的不是节省。:)
— 并不是说查尔斯（Charles）
@不是查尔斯：很好，完全错过了if a[v][u]<1冗余（因为非零单元格将具有min()的0）。
— Chas Brown '18
1

Clojure，247个字节

#(let[R(range(count %))C(range(count(% 0)))](loop[M % s 2](if-let[c(ffirst(sort-by last(for[x R y C :when(=((M x)y)0)][[x y](-(nth(sort(for[i R j C :when(>((M i)j)0)](+(*(- i x)(- i x))(*(- j y)(- j y)))))0))])))](recur(assoc-in M c s)(inc s))M)))

输入是VEC-的-血管内皮细胞M，它是在改进的loop通过assoc-in。如果找不到空闲点（if-let），则返回结果。

— 尼古妮
source
1

果冻，35 33 30 29字节

ZJæịþJFạþx¥F«/MḢṬ×FṀ‘Ɗo@FṁµÐL

在线尝试！

×ı+用æị（复杂的合并）替换了一个新的基于j.J 的对偶，从而节省了一个字节。

这是TIO的更有效版本。在线尝试！

说明

ZJæịþJFạþx¥F«/MḢṬ×FṀ‘Ɗo@FṁµÐL  Input: matrix M
Z                              Transpose
 J                             Enumerate indices - Get [1 .. # columns]
     J                         Enumerate indices - Get [1 .. # rows]
  æịþ                          Outer product using complex combine
                                 (multiply RHS by 1j and add to LHS)
      F                        Flatten
           F                   Flatten input
          ¥                    Dyadic chain
         x                       Times - Repeat each of LHS by each of RHS
       ạþ                        Outer product using absolute difference
            «/                 Reduce by minimum
              M                Indices of maximal values
               Ḣ               Head
                Ṭ              Untruth - Return a Boolean array with 1's at the indices
                 ×             Times
                     Ɗ         Monadic chain
                  F              Flatten input
                   Ṁ             Maximum
                    ‘            Increment
                      o@       Logical OR
                        F      Flatten input
                         ṁ     Mold - Reshape to match the input
                          µÐL  Repeat until result converges
— 英里
source
By using our site, you acknowledge that you have read and understand our Cookie Policy and Privacy Policy.
Licensed under cc by-sa 3.0 with attribution required.