Play on Words

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 N that indicates the number of
plates (1 N 100000). Then exactly N lines 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
32
acm
ibm
3
acm
malform
mouse
2
ok

ok

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

The door cannot be opened.

// 题意：输入n个单词，是否可以排成一个序列，使得每个单词的第一个字母和上一个单词的最后一个字母相同
// 算法：把字母看作结点，单词看成有向边，则有解当且仅当图中有欧拉路径。注意要先判连通
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;


const int maxn = 1000 + 5;


int pa[256];     //并查集
int findset(int x) { return pa[x] != x ? pa[x] = findset(pa[x]) : x; } 


int used[256], deg[256]; // 是否出现过；度数


int main() 
{
int T;
scanf("%d", &T);
while(T--) //测试数据组数;
{
int n;
char word[maxn];

scanf("%d", &n); //单词的个数;
memset(used, 0, sizeof(used));
memset(deg, 0, sizeof(deg));
for(int ch = 'a'; ch <= 'z'; ch++) pa[ch] = ch; // 初始化并查集,每个字母的祖先都是自己;
int cc = 26; // 连通块个数

for(int i = 0; i < n; i++)
{
scanf("%s", word); //接收字符串;
char c1 = word[0], c2 = word[strlen(word)-1]; //将首字母和最后一个字母取出;
deg[c1]++; //入度；
deg[c2]--; //出度;
used[c1] = used[c2] = 1; //记录字母出现的状态;
int s1 = findset(c1), s2 = findset(c2); //查找字母的祖先;
if(s1 != s2) { pa[s1] = s2; cc--; } //一个单词，首尾两个字母是连通的；
}

vector<int> d;
for(int ch = 'a'; ch <= 'z'; ch++) 
{
if(!used[ch]) cc--; // 没出现过的字母
else if(deg[ch] != 0) d.push_back(deg[ch]); //如果出度和入度不相同，压入数组；
}


bool ok = false;
if(cc == 1 && (d.empty() || (d.size() == 2 && (d[0] == 1 || d[0] == -1)))) ok = true;
if(ok) printf("Ordering is possible.\n");
else printf("The door cannot be opened.\n");
}
return 0;
}