n个不同数的组合

Question

1 2
3 1
2 4
3 2

 1 2
 | 
 3 1
  \
 2 4
   |
 3 2

 1 2
   | 
 3 1
   |
 2 4
  /
 3 2

我的问题是我无法思考如何编码，因为n<=50和a，b<=16，所以我不确定有多少个不同的数字，如果有16个数字，那么我必须找到16个数字的所有可能的组合，所以指导我通过这个。

Answer 1

要形成一个不同数字的列表，只需使用一个“唯一集”，并不断地将所有数字插入其中。在C++中，std::set by definition只存储唯一的数字。

为了找到不同序列的组合数量，您必须保留一个“候选列表”列表，如果他们已经没有这些数字，则继续在其中插入数字，否则删除特定的候选列表。

C++中的完整代码：

#include <iostream>
#include <vector>
#include <set>
using namespace std;

int main() {

    int n = 4;
    set<int> uniqueNumbers; // ordered set of unique numbers
    vector< set<int> > possibleLists( 1 );
    set<int>::iterator it;

    for ( int i = 0; i < n; i++ ) {

        int num1;
        int num2;
        cin >> num1 >> num2;

        // numbers will be inserted if not already present in set (by definition)
        uniqueNumbers.insert( num1 );
        uniqueNumbers.insert( num2 );

        // make a copy for a possible new branch
        vector< set<int> > possibleListsCopy( possibleLists );

        //int size1 = possibleLists.size();

        for ( int j = 0; j < possibleLists.size(); j++ ) {

            it = possibleLists[j].find( num1 );
            if ( it == possibleLists[j].end() ) {
                possibleLists[j].insert( num1 ); // insert if not found
                //cout << "inserted1 "<<endl;
            }
            else {
                // erase this possible combination
                possibleLists[j].clear();
                possibleLists.erase( possibleLists.begin() + j );
                j--;
            }

        }

        //int size2 = possibleListsCopy.size();

        for ( int j = 0; j < possibleListsCopy.size(); j++ ) {
;

            it = possibleListsCopy[j].find( num2 );
            if ( it == possibleListsCopy[j].end() ) {
                possibleListsCopy[j].insert( num2 ); // insert if not found
            }
            else {
                // erase this possible combination
                possibleListsCopy[j].clear();
                possibleListsCopy.erase( possibleListsCopy.begin() + j );
                j--;
            }

        }

        // concatenate both set of lists.
        possibleLists.insert( possibleLists.end(),
                                possibleListsCopy.begin(),
                                possibleListsCopy.end() );


    }


    cout << " The unique list: ";
    //output the unique list.
    for ( it = uniqueNumbers.begin(); it != uniqueNumbers.end(); it++ )
        cout << *it << " ";

    /*cout << endl << endl;

    cout << "Possible Lists:" << endl;

    for ( int i = 0; i < possibleLists.size(); i++ ) {

        for ( it = possibleLists[i].begin(); it != possibleLists[i].end(); it++ )
            cout << *it << " ";
        cout << endl;

    }*/

    cout << endl << "Total number of combinations: "
        << possibleLists.size() << endl;

    return 0;
}