样例过了求hack

Problem D: 单词游戏
Time Limit: 1 Sec  Memory Limit: 32 MB
Description
有 N个盘子，每个盘子上写着一个仅由小写字母组成的英文单词。你需要给这些盘子安排一个合适的顺序，使得相邻两个盘子中，前一个盘子上单词的末字母等于后一个盘子上单词的首字母。请你编写一个程序，判断是否能达到这一要求。如果能，请给出一个合适的顺序。
Input
多组数据。第一行给出数据组数 T，每组数据第一行给出盘子数量 N，接下去 N 行给出小写字母字符串，一种字符串可能出现多次。
Output
若存在一组合法解输出"Ordering is possible."，否则输出"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.

Sample Input2
5
2
acb
bac
2
bac
acb
2
abc
def
3
abc
cba
cba
4
abc
cba
abc
abc
Sample Output2
Ordering is possible.
Ordering is possible.
The door cannot be opened.
Ordering is possible.
The door cannot be opened.

HINT
1≤N≤10^5,∣S∣≤1000

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define M 200001
#define can "Ordering is possible."
#define cant "The door cannot be opened."
#define clean(a) memset(a, 0, sizeof(a))
using namespace std;
struct E { int to, next; } e[M];
char s[1001];
int T, n, u, v, st, dif_1, dif1;
int ins[26], outs[26], head[26], tot;
bool exist[26], vis[26], pd;
void adde(int x, int y)
{
	e[++tot].next = head[x];
	head[x] = tot;
	e[tot].to = y;
}
void dfs(int u)
{
	vis[u] = true;
	for(int i=head[u]; i; i=e[i].next)
	{
		int v = e[i].to;
		if(!vis[v])
			dfs(v);
	}
}
int main()
{
	scanf("%d", &T);
	while(T--)
	{
		tot = dif_1 = dif1 = 0;
		pd = true;
		clean(ins);
		clean(outs);
		clean(head);
		clean(exist);
		clean(vis);
		
		scanf("%d", &n);
		if(n==1)
		{
			scanf("%s", s);
			puts(can);
			continue;
		}
		
		for(int i=1; i<=n; i++)
		{
			scanf("%s", s);
			u = s[0]-'a';
			v = s[strlen(s)-1]-'a';
			outs[u]++;	ins[v]++;
			exist[u] = exist[v] = true;
			adde(u, v), adde(v, u);
		}
		
		for(int i=0; i<26; i++)
		{
			int dif = ins[i]-outs[i];
			if(abs(dif)>1)
			{
				pd = false;
				break;
			}
			else if(dif == -1)	dif_1 ++, st = i;
			else if(dif == 1)	dif1 ++;
		}
		if(!pd || dif_1>1 || dif1>1 || (dif_1 ^ dif1)==1)
		{
			puts(cant);
			continue;
		}
		
		dfs(st);
		for(int i=0; i<26; i++)
		{
			if(exist[i] && !vis[i])
			{
				pd = false;
				break;
			}
		}
		
		if(pd)	puts(can);
		else	puts(cant);
	}
	
	return 0; 
}