題意：判斷所有字串是不是可以連在一起，實際上就是判斷所構造的有向圖是否存在歐拉回路或者尤拉圖；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include <algorithm>
#define N 100010
using namespace std;
char c[N][2];
int al[123];
int in[27],out[27];
int f[27],ccount;
bool yo[26];
int tou(int x)
{
    if(f[x]!=x)
    {
        f[x]=tou(f[x]);
    }
    return f[x];
}
void build(int x,int y)
{
    int toux=tou(x),touy=tou(y);
    if(touy!=toux)
    {
        ++ccount;
        f[toux]=touy;
    }
}

int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        memset(al,0,sizeof(al));
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(yo,0,sizeof(yo));
        memset(f,0,sizeof(f));


        scanf("%d%*c",&n);
        int geshu=0;
        for(int i=0; i<n; i++)
        {
            char cc,ccc;
            scanf("%c",&c[i][0]);
            if(!yo[c[i][0]-'a'])
            {
                yo[c[i][0]-'a']=1;
                ++geshu;
            }
            while(scanf("%c",&cc)&&cc!='\n')
            {
                ccc=cc;
            }
            c[i][1]=ccc;
            if(!yo[c[i][1]-'a'])
            {
                yo[c[i][1]-'a']=1;
                ++geshu;
            }
            in[c[i][1]-'a']++;
            out[c[i][0]-'a']++;
        }
        for(int i=0; i<26; i++)
            f[i]=i;

        ccount=0;
        for(int i=0; i<n; i++)
            build(c[i][0]-'a',c[i][1]-'a');
        int count=0,flag=1;

        if(ccount+1!=geshu)
            flag=0;
        if(flag)
            for(int i=0; i<26; i++)
            {
                if(in[i]!=out[i])
                {
                    count++;
                    if(abs(in[i]-out[i])!=1||count>2)
                    {
                        flag=0;
                        break;
                    }
                }
            }

        if(!flag)
            printf("The door cannot be opened.\n");
        else
            printf("Ordering is possible.\n");
    }
    return 0;
}