Play on Words(欧拉回路)

Problem Description
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 wordmotorola’’. 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
题意就是给出一些单词，问是否能组成一个序列，让序列的前一个单词的尾字母与后一个单词的头字母相同；类似于成语接龙。
是否组成欧拉回路：1.判断图是否连通 2.该图所有顶点的出度等于入度或者 一个顶点的度数为1，另一个度数为-1，其他顶点的出度等于入度。
代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
const int maxn=100;
int n;
int G[maxn][maxn],vis[maxn],in[maxn],out[maxn];
void dfs(int u)
{
	vis[u]=1;        //访问标记 
	for(int v=0;v<26;v++)     //字母节点
	if(!vis[v]&&G[u][v])
	dfs(v);           //没有被访问并且有有向边
}


int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{		
		char s[100005];
		memset(vis,1,sizeof(vis));       //标志是否被遍历过
		memset(G,0,sizeof(G));
		memset(in,0,sizeof(in));         //入度
		memset(out,0,sizeof(out));       //出度
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%s",s);
			int len=strlen(s);
			int r=s[0]-'a';       //首 
			int t=s[len-1]-'a';    //尾 
			G[r][t]=1;              //有有向边
			in[t]++;
			out[r]++;          //很重要！！！，出入度别写反了,
			vis[r]=vis[t]=0;         //将其置为未访问状态
		}
		int flag=0,cnt1=0,cnt2=0,cnt3=0,p=0;
		for(int i=0;i<26;i++)
		{
			if(in[i]==out[i])        //环的情况
			continue;
			else if(out[i]==in[i]+1)     //起点 
			{
				p=i;
				cnt1++;
			}
			else if(in[i]==out[i]+1)      //终点 
			cnt2++;
			else cnt3++;   //没有满足
		}
		if(cnt3>0) 
		{
			printf("The door cannot be opened.\n");
			continue;
		}
		if(cnt1==0&&cnt2==0 || cnt1==1&&cnt2==1)      //组成环或者一条直线 
		flag=1;
		else flag=0;
		dfs(p);
		for(int i=0;i<26;i++)
		if(!vis[i]) flag=0;
		if(flag==1) printf("Ordering is possible.\n");
		else printf("The door cannot be opened.\n");
	}
	return 0;
}