In your childhood you most likely had to solve the riddle of the house of Santa Claus. Do you remember that the importance was on drawing the house in a stretch without lifting the pencil and not drawing a line twice? As a reminder it has to look like shown in Figure 1.

Figure: The House of Santa Claus

Well, a couple of years later, like now, you have to ``draw'' the house
again but on the computer. As one possibility is not enough, we require
all the possibilities when starting in the lower left corner. Follow the example in Figure 2 while defining your stretch.

Figure: This Sequence would give the Outputline 153125432

All the possibilities have to be listed in the outputfile by increasing order, meaning that 1234... is listed before 1235... .

Output

So, an outputfile could look like this:

12435123

13245123

...

1512342

分析：
用5个点表示圣诞老人的房子，除了1-4和2-4两条边外，所有的边都相连，问能否从左下角（点1）一笔画出房子的路径有哪些，逐个点输出出来。
例如：
12435123
13245123
......
15123421
分析：
从点1开始，深搜遍历所有的边，直到遍历的边的个数达到8时递归结束。
AC code:

#include<bits/stdc++.h>

using namespace std;

int m[][];

void init()

{

    for(int i=;i<=;i++)

    {

        for(int j=;j<=;j++)

        {

            if(i!=j)    m[i][j]=;

            else m[i][j]=;

        }

    }

    m[][]=m[][]=;

    m[][]=m[][]=;

}

void dfs(int e,int k,string s)

{

    s+=(k+'');

    if(e==)

    {

        cout<<s<<endl;

        return;

    }

    for(int i=;i<=;i++)

    {

        if(m[k][i])

        {

            m[i][k]=m[k][i]=;

            dfs(e+,i,s);

            m[i][k]=m[k][i]=;

        }

    }

}

int main()

{

    init();

    dfs(,,"");

}