Two-way recursive algorithm problem

First question: Give the stacking order of {1, 2, 3,..., n}, and output all possible stacking order

#include"stdafx.h"#include<stack>#include<queue>#include<iostream>#include<cstdio>#include<cstdlib>using namespacestd;intn =0; typedef stack<int>stack;typedef Queue<int>Queue;voidDfsintLevel,stack s, Queue buf,Const intLevel );/** Each recursive has only one element (level) in the stack, but there may be 0 elements out of the stack, there may be at least one element out of the stack * @param level present scale the size of current Sub-problem * @param s current sub-problem stack the stack of current Sub-problem * @param buf the print queue of the present sub-issue the printing queue of the Sub-p Roblem * @param level largest The maximum scale of the problem*/voidDfsintLevel,stack s, Queue buf,Const intLevel ) {    //by using ' #pragma region xxx ' direction, the code can appear more gracefully~#pragmaRegion termination of recursionif(Level = =Level ) {        //First We print all elements in the queueBuf.push (level);  while(!Buf.empty ()) {cout<<buf.front () <<" ";        Buf.pop (); }        //Then we clear the stack         while(!S.empty ()) {cout<< s.top () <<" ";        S.pop (); } cout<<Endl; N++; return; }#pragmaEndregion, when Level==level, the recursion stops.Queue Tempqueue (BUF);    S.push (level);    Stack Tempstack (s); /** Very Important steps * Break the current problem into several sub-problems * must take into account all possible sub-problems, such as * 1. Push then no pop * 2. Push then pop 1 * 3. Push then pop 2 * ...*/     while(!S.empty ())        {Buf.push (S.top ());        S.pop (); DFS ( level+1, S, buf, level); } DFS ( level+1, Tempstack, tempqueue, level);}int_tmain (intARGC, _tchar*argv[]) {unsignedintcases; intx; scanf ("%d", &cases);  while(cases--) {Stack s=Stack (); Queue Q=Queue (); CIN>>x; DFS (1, S, q, x); cout<<"The Catalan number of"<<x<<" is"<<n<<Endl; N=0; } System ("shutdown-s-T");}

Two.

Input a positive integer n (n>=3), write a function to print all the possibilities of combination of 1,1,2,2,3,3,......, n , n (2n numbers altogether) that satisfies:
There is 1 integer between "1" s
There is 2 integers between "2" s
There is 3 integers between "3" s
......
There is n integers between "n" s
For example of 3,the, possible sequence are:231213 or 312132.
Pay attention, there is 0 or many possible sequences, print them all

Result class Result.h

#pragmaOnceclassresult{Private:    int*data; intN; Public: Result (int_n): N (_n) {data=New int[2*N];    Recover (); }    ~Result () {delete[] data;} Result (Constresult&src) {N=SRC.N; Data=New int[2*N];  for(intI=0; i<2*n; ++i) {Data[i]=Src[i]; }    }    int&operator[](intindex) {        returnData[index]; }    Const int&operator[] (intIndexConst    {        returnData[index]; }    BOOLInsertintIintLevel ) {        if(i<0|| i+level+1>2*n)return false; if(data[i]==0&& data[i+level+1]==0) {Data[i]= data[i+level+1] =Level ; return true; }        Else return false; }    voidrecover () { for(intI=0; i<2*n; ++i) {Data[i]=0; }} Result&operator=(Constresult&src)        {delete[] data; N=SRC.N; Data=New int[2*N];  for(intI=0; i<2*n; ++i) {Data[i]=Src[i]; }        return* This; }    intSize ()Const    {        return 2*N; }    voidPrint ()Const    {         for(intI=0; i<2*n; ++i) printf_s ("%d", Data[i]); printf_s ("\ n"); }};

Core algorithms:

#include"stdafx.h"#include"Result.h"#include<stack>#include<queue>#include<list>#include<vector>using namespacestd;voidDfsintLevel , Result Res) {    if(level<1) {res.print (); return; } Result Temp=Res;  for(intI=0; I<res.size ()-level-1; ++i) {if(Temp.insert (I, level)) {DFS ( level-1, temp); Temp=Res; }        Else Continue; }}int_tmain (intARGC, _tchar*argv) {    intcases; scanf_s ("%d", &cases);  while(cases--)    {        intN; scanf_s ("%d", &N); Result Res=Result (n);    DFS (n, res); }}

Two-way recursive algorithm problem