所有不相交对的集合 [英] Sets of all disjoint pairs
问题描述
给定一组 {1,2,3,4,5...n}
的 n
个元素,我们需要找到所有不相交对的集合.
Given a set {1,2,3,4,5...n}
of n
elements, we need to find all sets of disjoint pairs.
例如,如果 n=4,则输出为
For example, if n=4, the output would be
{(1,2),(3,4)}, {(1,3),(2,4)}, {(1,4),(2,3)}
我什至不知道如何开始.我希望有人能给我一个关于使用哪种算法的建议,可能还有一些实现细节.
I am not even able to figure out how to start. I am hoping someone can give me a suggestion about which algorithm to use, and possibly some implementation details as well.
推荐答案
递归生成(n-1)的Delphi代码!!从 n=2*k 个元素中设置 (1*3*5*7...n-1)
Delphi code for recursive generation of (n-1)!! sets (1*3*5*7...n-1) from n=2*k elements
var
A: TArray<Integer>;
procedure Swap(i, j: integer);
var
t : integer;
begin
t := A[i];
A[i] := A[j];
A[j] := t;
end;
procedure MakePairs(Start: Integer; Pairs: string);
var
i: Integer;
begin
if Start >= Length(A) then
Writeln(Pairs)
else
for i := Start + 1 to High(A) do begin
Swap(Start + 1, i); //store used element in the array beginning
MakePairs(Start + 2, Pairs + Format('(%d,%d)', [A[Start], A[Start + 1]]));
Swap(Start + 1, i); //get it back
end;
end;
begin
A := TArray<Integer>.Create(1,2,3,4,5,6);
//be sure that array length is even!!!
MakePairs(0, '');
Writeln(PairCount);
输出:
(1,2)(3,4)(5,6)
(1,2)(3,5)(4,6)
(1,2)(3,6)(5,4)
(1,3)(2,4)(5,6)
(1,3)(2,5)(4,6)
(1,3)(2,6)(5,4)
(1,4)(3,2)(5,6)
(1,4)(3,5)(2,6)
(1,4)(3,6)(5,2)
(1,5)(3,4)(2,6)
(1,5)(3,2)(4,6)
(1,5)(3,6)(2,4)
(1,6)(3,4)(5,2)
(1,6)(3,5)(4,2)
(1,6)(3,2)(5,4)
15
加法
也适用于奇数长度数组的变体(奇怪的排序)
Addition
Variant that works with odd-length array too (weird ordering)
procedure MakePairs(Start: Integer; Pairs: string);
var
i: Integer;
OddFlag: Integer;
begin
if Start >= Length(A) then
Memo1.Lines.Add(Pairs)
else begin
Oddflag := (High(A) - Start) and 1;
for i := Start + OddFlag to High(A) do begin
Swap(Start + OddFlag, i);
if OddFlag = 1 then
MakePairs(Start + 2, Pairs + Format('(%d,%d)', [A[Start], A[Start + 1]]))
else
MakePairs(Start + 1, Pairs);
Swap(Start + OddFlag, i);
end;
end;
end;
对于 (1,2,3,4,5):
for (1,2,3,4,5):
(2,3)(4,5)
(2,4)(3,5)
(2,5)(4,3)
(1,3)(4,5)
(1,4)(3,5)
(1,5)(4,3)
(2,1)(4,5)
(2,4)(1,5)
(2,5)(4,1)
(2,3)(1,5)
(2,1)(3,5)
(2,5)(1,3)
(2,3)(4,1)
(2,4)(3,1)
(2,1)(4,3)
15
现在不相关:
如果每一对只出现一次(从你的例子中 n=4 不清楚),那么你可以使用 循环赛算法
Not relevant now:
If every pair should occur just once (it is not clear from your example with n=4), then you can use round-robin tournament algorithm
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