Some of the secret doors contain a very interesting word puzzle. The team of archaeologists has to solve it to open that doors. Because there is no other way to open the doors, the puzzle is very important for us.

There is a large number of magnetic plates on every door. Every
plate has one word written on it. The plates must be arranged into
a sequence in such a way that every word begins with the same
letter as the previous word ends. For example, the word ``acm'' can
be followed by the word ``motorola''. Your task is to write a
computer program that will read the list of words and determine
whether it is possible to arrange all of the plates in a sequence
(according to the given rule) and consequently to open the
door.

The input
consists of T test cases. The number of them (T) is given on the
first line of the input file. Each test case begins with a line
containing a single integer number Nthat indicates the number of
plates (1 <= N <= 100000). Then
exactly Nlines follow, each containing a single word. Each word
contains at least two and at most 1000 lowercase characters, that
means only letters 'a' through 'z' will appear in the word. The
same word may appear several times in the list.

Your program
has to determine whether it is possible to arrange all the plates
in a sequence such that the first letter of each word is equal to
the last letter of the previous word. All the plates from the list
must be used, each exactly once. The words mentioned several times
must be used that number of times.
If there exists such an ordering of plates, your program should
print the sentence "Ordering is possible.". Otherwise, output the
sentence "The door cannot be opened.".

那么就是 a->m  m->m
m->e  
=>  
a->m->e
再来看看这样算不合法的
o->k
o->k
就不合法了，o的出度为2，入度为0，k的入度为2，出度为0
a->m
i->m
也不合法，m的入度为2，出度为0

该问题可以转换为欧拉回路问题。
1，保证改图是连通的。
2，如果它是环形，那么它的每一个结点的出度=入度。
3，如果它是链状，那么只有它的起始结点（出度-入度=1）和结束结点（入度-出度=1）

#include <stdio.h>

#include <string.h>

int a[30];

int flag[30];

int abs(int a){

    if(a<0)return -a;

    else return a;

}

int find(int x){

    int temp=a[x];

    while(temp!=a[temp]){

        temp=a[temp];

    }

    int v;

    while(x!=a[x]){

        v=a[x];

        a[x]=temp;

        x=v;

    }

    return temp;

}

void u(int x,int y){

    x=find(x);

    y=find(y);

    if(x!=y){

        a[x]=y;

    }

}

int main(int argc, char *argv[])

{

    int t;

    int begin,end;

    char ch[1001];

    scanf("%d",&t);

    while(t--){

        int n;

        int in[30],out[30];

        scanf("%d",&n);

        memset(flag,0,sizeof(flag));

        memset(in,0,sizeof(in));

        memset(out,0,sizeof(out));

        for(int i=0; i<=25; i++)a[i]=i;

        while(n--){

            scanf("%s",ch);

            int len=strlen(ch);

            begin=ch[0]-'a';

            end=ch[len-1]-'a';

            flag[begin]=1;

            flag[end]=1;

            in[end]++;

            out[begin]++;

            u(begin,end);

        }

        int s=0;

        for(int i=0; i<=25; i++){

            if(flag[i]==1 && a[i]==i)s+=1;

        }

        if(s==1){

            int f_in=0,f_out=0;

            int k;

            for( k=0; k<=25; k++){

                if(flag[k]==1){

                    if(in[k]==out[k]){

                        continue;

                    }else if(abs(in[k]-out[k])==1){

                        if(in[k]-out[k]<0)f_out++;

                        if(in[k]-out[k]>0)f_in++;

                        if(f_out>1 || f_in>1)break;

                    }else if(abs(in[k]-out[k])>1){

                        break;

                    }

                }

            }

            if(k==26){

                puts("Ordering is possible.");

            }else{

                puts("The door cannot be opened.");

            }

        }else{

            puts("The door cannot be opened.");

        }

    }

    return 0;

}