生成{0,1,2,... n-1}的所有大小k个子集, [英] Generating all size k subsets of {0, 1, 2, ... n-1}
问题描述
我想生成所有基数 k
{0,1,2,...,n-1}的子集
在C ++。在Haskell中,我会:
sets 0 n = [[]]
设置kn = [i:s | i < - [0..n-1],s < - set(k-1)i]
$ b b
或在Python中:
def sets(k,n):
if k == 0:
return [()]
return((i,)+ s for(in)(n)for s in sets(k-1,i))
例如,(为了清楚起见,换行符)
ghci>设置2 8
[[1,0],
[2,0],[2,1],
[3,0],[3,1],[3,2 ],
[4,0],[4,1],[4,2],[4,3],
[5,0],[5,1] ],[5,3],[5,4],
[6,0],[6,1],[6,2],[6,3],[6,4] ,5],
[7,0],[7,1],[7,2],[7,3],[7,4],[7,5],[7,6]
这样做的C ++方式是什么?请注意,我不是要求如何来解决这个问题。我想知道什么数据类型被C ++程序员认为是正常的。
(参考,我熟悉C ++, 。
解决方案这里是一个朴素的递归方法,它实现了古典的组合身份:
binom(n + 1,k + 1)= binom(n,k + 1)+ binom b
#include< set>
typedef std :: set< int> intset;
std :: set< intset>子集(std :: size_t k,intset s)
{
if(k == 0 || s.empty()|| s.size()< k){return {{} ; }
if(s.size()== k){return {s}; }
auto x = * s.begin();
s.erase(s.begin());
std :: set< intset>结果;
for(auto& t:subsets(k - 1,s))
{
auto r = std :: move(t);
r.insert(x);
result.insert(std :: move(r));
}
for(auto& t:subsets(k,s))
{
results.insert(std :: move(t));
}
返回结果;
}
用法:
auto ss = subsets(3,{0,1,2,3,4});
完成示例:
#include< iostream>
#include< string>
#include< prettyprint.hpp>
int main(int argc,char * argv [])
{
if(argc!= 3)return 1;
auto k = std :: stoul(argv [1]);
auto n = std :: stoul(argv [2]);
intset s;
for(auto i = 0U; i!= n; ++ i)s.insert(i);
std :: cout<<子集(k,s) std :: endl;
}
I want to generate all cardinality
k
subsets of{0, 1, 2, ..., n-1}
in C++. In Haskell, I would do:sets 0 n = [[]] sets k n = [i:s | i <- [0..n-1], s <- sets (k-1) i]
Or in Python:
def sets(k, n): if k == 0: return [()] return ((i,)+s for i in range(n) for s in sets(k-1, i))
So, for example, (line breaks added for clarity)
ghci> sets 2 8 [[1,0], [2,0],[2,1], [3,0],[3,1],[3,2], [4,0],[4,1],[4,2],[4,3], [5,0],[5,1],[5,2],[5,3],[5,4], [6,0],[6,1],[6,2],[6,3],[6,4],[6,5], [7,0],[7,1],[7,2],[7,3],[7,4],[7,5],[7,6]]
What would be the "C++ way" of doing this? Note that I'm not asking how to solve the problem. I'm asking about what data types would be considered "normal" by C++ programmers.
(For reference, I'm vaguely familiar with C++ and somewhat familiar with C.)
解决方案Here's a naive, recursive approach, which implements the classical combinatorial identity:
binom(n + 1, k + 1) = binom(n, k + 1) + binom(n, k)
#include <set> typedef std::set<int> intset; std::set<intset> subsets(std::size_t k, intset s) { if (k == 0 || s.empty() || s.size() < k) { return { { } }; } if (s.size() == k) { return { s }; } auto x = *s.begin(); s.erase(s.begin()); std::set<intset> result; for (auto & t : subsets(k - 1, s)) { auto r = std::move(t); r.insert(x); result.insert(std::move(r)); } for (auto & t : subsets(k, s)) { results.insert(std::move(t)); } return result; }
Usage:
auto ss = subsets(3, {0, 1, 2, 3, 4});
Complete example:
#include <iostream> #include <string> #include <prettyprint.hpp> int main(int argc, char * argv[]) { if (argc != 3) return 1; auto k = std::stoul(argv[1]); auto n = std::stoul(argv[2]); intset s; for (auto i = 0U; i != n; ++i) s.insert(i); std::cout << subsets(k, s) << std::endl; }
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