如何有效地生成组合而不重复它们之间具有某些特殊数字

Question

如何在不重复的情况下有效地生成数字组合的集合,其中所有集合在彼此之间具有特定的独特数字.
*注意:范围编号始终从0开始.

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示例:

范围编号(数字[]) = 0,1,2,3,4,5,6,7 ==>总共8个数字(n).
组合(k) = 5个数字.
特殊数字(nD) = 2个数字.

结果:
0 1 2 3 4
0 1 2 5 6
0 1 3 5 7
0 1 4 6 7
0 2 3 6 7
0 2 4 5 7
0 3 4 5 6
共有7种有效组合

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如何组装:

由于我不善于言语,所以让我把它们想象成这样:

要解释他们的独特数字:

我们可以将它们总结到这个表中:

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到目前为止我取得了什么成就

我目前的解决方案效率很低(或者你可以称之为暴力).
*首先我为每个组合循环.==> k C n
*然后我为有效组合创建一个temp.
*然后对于每个组合我验证我的临时,如果它有效然后将其存储在临时,否则忽略它.
而已.

这是我在Console App中的代码:

class Program
{
    static List<int[]> ValidCombinations;

    static void Main()
    {
        ValidCombinations = new List<int[]>();

        int[] numbers = Enumerable.Range(0, 8).ToArray();
        int n = numbers.Length;
        const int k = 5;
        const int nD = 2;

        int maxIntersect = k - nD;

        int iCombination = 0;
        int iValidCombination = 0;
        int[] _temp = new int[k];
        foreach (int[] c in FindCombinations(k, n))
        {
            // #Print out
            for (int i = 0; i < n; i++)
            {
                if (c.Contains(i))
                    Console.Write(c[Array.IndexOf(c, i)] + " ");
                else
                    Console.Write("_ ");
            }

            // Save to List
            if (IsValidSet(c, maxIntersect))
            {
                _temp = new int[k];
                for (int i = 0; i < c.Length; i++)
                {
                    _temp[i] = c[i];
                }
                ValidCombinations.Add(_temp);
                iValidCombination++;
                Console.Write(" ### --> {0}", string.Join(" ", c));
            }
            Console.WriteLine();

            iCombination++;
        }
        Console.WriteLine("\nTotal Combination = {0}", iCombination);
        Console.WriteLine("Valid Combination Found = {0}", iValidCombination);
    }

    public static IEnumerable<int[]> FindCombosRec(int[] buffer, int done, int begin, int end)
    {
        for (int i = begin; i < end; i++)
        {
            buffer[done] = i;

            if (done == buffer.Length - 1)
                yield return buffer;
            else
                foreach (int[] child in FindCombosRec(buffer, done + 1, i + 1, end))
                    yield return child;
        }
    }

    public static IEnumerable<int[]> FindCombinations(int m, int n)
    {
        return FindCombosRec(new int[m], 0, 0, n);
    }

    private static bool IsValidSet(int[] set, int maxIntersect)
    {
        foreach (var item in ValidCombinations)
        {
            if (set.Intersect(item).Count() > maxIntersect)
                return false;
        }

        return true;
    }
}

我从这里得到了基本代码来生成组合.

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问题

这是有效的,但对于更大范围的数字,此解决方案将花费大量时间来完成.我知道因为所涉及的组合算法,但必须有某种捷径或模式来简化它(我的小脑子没能弄明白).

非常感谢你.

Answer 1

#include<iostream>
#include<vector>
#define N 8
#define K 5
#define D 2
using namespace std;

vector<vector<int>> vv;
vector<int> v;
int intersection(const vector<int>& a, const vector<int>& b) 
{//count elements of intersection of two sorted vectors
    int count = 0;
    auto a_it = a.begin();
    auto b_it = b.begin();
    while(a_it != a.end() && b_it != b.end()) {
        if(*a_it == *b_it) count++, a_it++, b_it++; 
        else if(*a_it < *b_it) a_it++;
        else b_it++;
    }
    return count;
}

void select_num(int n)
{//might reduce some unnecessary iteration of nCk combination 
    for(auto& a : vv) if(intersection(a, v) > K - D) return;
    //above line will cut off the chain when the intersection is already over 
    //limit. You can add some more conditions to cut off unnecessary calculation.
    if(v.size() == K) {
        bool ok = true;
        for(auto& a : vv) {
            if(intersection(a, v) != K - D) {
                ok = false;
                break;
            }
        }
        if(ok) vv.push_back(v);
        return;
    }
    if(n == N) return;

    //case : select n
    v.push_back(n);
    select_num(n+1);
    v.pop_back();

    //case : do not select n
    select_num(n+1);
}

int main()
{
    select_num(0);
    for(auto& a : vv) {
        for(auto& b : a) cout << b << ' ';
        cout << endl;
    }
    cout << endl << vv.size() << endl;
}