Play on Words HDU - 1116 （并查集+欧拉通路）

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.

大佬的题解：

这是一道欧拉回路的简单题目，对于初学者的理解很有帮助，所以把它分享一下。这是一个有向图。

这道题大意可抽象为把26个字母个字母看做点（所以这里我们需要把字母与点进行转化），然后把一些点联通到一起，判断最后可不可以组成线或者环。

组成线的条件为所有点都将通过其他点链接到一个父亲节点上，而组成环的条件为所有点都与其他点相连并且出度=入度，只有两个点入度与出度差1,。

题中涉及一些并差集相关知识，来计算图中有几条线。

注：所有点必须用上。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
int f[30],in[30],out[30];
int getf(int v)
{
	while(v!=f[v])
	v=f[v];
	return v;
}
void merge(int u,int v)
{
	int t1=getf(u);
	int t2=getf(v);
	f[t2]=t1;
	return;
}

int main(void)
{
	int t;
	cin>>t;
	while(t--)
	{
		int n;
		char a[1010];
		int u,v;
		for(int i=0;i<27;i++)
		{
			f[i]=i;
			in[i]=0;
			out[i]=0;
		}
		scanf("%d",&n);
		while(n--)
		{
			scanf("%s",a);
			v=a[0]-'a';
			u=a[strlen(a)-1]-'a';
			in[v]++;  //入度 
			out[u]++; //出度 
			merge(v,u);
		}
		int f1=0;;//记录入度数不等于出度数的点的个数，即非联通点的个数
		int f2=0;//记录点联通后有几条线，即全部点是否联通 
		int f3=0;//记录线中点的个数
		n=0;   // 记录出现点数 
		for(int i=0;i<27;i++)
		{
			if(in[i]==0&&out[i]==0)continue; //如果点没出现，不记录点的个数
			n++;
			if(abs(in[i]-out[i])==1)f1++;
			if(f[i]==i)f2++;
			if(in[i]==out[i])f3++;
		}
		if(f2==1&&(f3==n||(f1==2&&f3==n-2)))
		printf("Ordering is possible.\n");
		else
		printf("The door cannot be opened.\n");
	}
	return 0;
 }