题目解析：

输入一些英文单词，根据该单词的首尾字母，判断所有单词能不能连成一串， 类似于成语接龙的意思。同样如果有多个重复的单词时，也必须满足这样的条件才能通过， 否则都是不可能的情况。输入包括若干个案例，每个案例中最多有100000个单词。

题目：

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. 

Input

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. 

Output

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.". 

Sample Input

3
2
acm
ibm
3
acm
malform
mouse
2
ok
ok

Sample Output

The door cannot be opened.
Ordering is possible.
The door cannot be opened.

思路解析：

该题所涉及的知识点主要是并查集和欧拉回路，欧拉回路又是什么呢？它跟我们 今天所做的“成语接龙”有什么联系呢？学过离散的同学都知道，欧拉回路是指，在图中，经 过所有的边一次且仅一次的回路就是欧拉回路。而经过所有的顶点一次且仅一次的回路叫做 哈密顿回路。

现在我们考虑每个单词有用的成分只有首尾字母，那么我们把单词的首尾字母 提取出来，抽象成两个顶点，而恰恰这两个顶点又是有了联系的，我们把它们放入一个集合 。实际上，无论输入多少个单词，首尾字母中所出现的字母无外乎a,b,c……y,z等26个英文字 母。所以反客为主，我们不用被动的根据输入的单词来变换，首先建立26个集合，将有联系 的集合合并（merge）则矣。当做完这个工作后，我们再来看看欧拉回路存在性的判定定理：

一、无向图 每个顶点的度数都是偶数，则存在欧拉回路。

二、有向图（所有边都是单向的） 每个节顶点的入度都等于出度，则存在欧拉回路。

 三.混合图欧拉回路 混合图欧拉回路用的是网络流。

把该图的无向边随便定向，计算每个点的入度和出度。如果有某个点出入度之差为奇数，那 么肯定不存在欧拉回路。因为欧拉回路要求每点入度 = 出度，也就是总度数为偶数，存在 奇数度点必不能有欧拉回路。

AC代码：

#include<iostream>
#include<string.h>
#define inf 0x3f3f3f3f
using namespace std;
int t,m,book[110];
char w[1010];
int  dp[110],u[110],v[110];
int gets(int xx)
{if(xx==dp[xx])return xx;elsereturn gets(dp[xx]);
}
void dfs(int x,int y)
{int dx=gets(x);int dy=gets(y);if(dx!=dy)dp[dy]=dx;
}
int main()
{cin>>t;while(t--){for(int i=0; i<26; i++)dp[i]=i;cin>>m;memset(book,0,sizeof(book));memset(u,0,sizeof(u));memset(v,0,sizeof(v));/*care*/for(int i=0; i<m; i++){cin>>w;int l=strlen(w);int a=w[0]-'a';int b=w[l-1]-'a';book[a]=1;book[b]=1;u[a]++;v[b]++;dfs(a,b);}int da=0,db=0;int flag=1;for(int i=0; i<26; i++){if(!flag)break;if(!book[i])continue;if(u[i]-v[i]>1||u[i]-v[i]<-1)flag=0;if(u[i]-v[i]==1){da++;if(da>1)flag=0;}if(u[i]-v[i]==-1){db++;if(db>1)flag=0;}}int ans=0;for(int i=0; i<26; i++)if(book[i]&&dp[i]==i){ans++;if(ans>1){flag=0;break;}}if(!flag)cout<<"The door cannot be opened."<<endl;elsecout<<"Ordering is possible."<<endl;}return 0;
}