A - Play on Words

题目描述:
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
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.

题目大意：
输入若干组单词,如果a单词末尾字母与b单词首字母相同,则a和b可以相接,问这若干组是否可以连成一条线。

思路：
类似于小时候玩的一笔画问题,
我们把输入的每一组单词看成图的一个边,首尾字母看成是一个点,此题即是一道判断欧拉通路的问题了,因为需要判断联通,可以用并查集解决。

AC代码:

#include<iostream>
#include<string.h>
using namespace std;

int enter,n,book[110];
char code[1010];
int  dp[110],u[110],v[110];

int find(int x);
void dfs(int x,int y);

int main() 
{
	cin >> enter;
	while(enter --) 
	{
		for(int i = 0; i < 26; i ++)
		{
			dp[i] = i;
		}
			
		cin >> n;
		
		memset(book,0,sizeof(book));
		memset(u,0,sizeof(u));
		memset(v,0,sizeof(v));
		
		for(int i = 0; i < n; i ++) 
		{
			cin >> code;
			int l = strlen(code);
			int a = code[0] - 'a';
			int b = code[l-1] - 'a';
			book[a] = 1;
			book[b] = 1;
			u[a] ++;
			v[b] ++;
			dfs(a,b);
		}
		
		int a = 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) 
			{
				a ++;
				
				if(a > 1)
				{
					flag = 0;
				}	
			}
			
			if(u[i] - v[i] == -1) 
			{
				db ++;
				
				if(db > 1)
					flag = 0;
			}
		}
		
		int xx = 0;
		for(int i = 0; i < 26; i ++)
			if(book[i] && dp[i] == i) 
			{
				xx ++;
				if(xx > 1) 
				{
					flag = 0;
					break;
				}
			}
			
		if(!flag)
			cout << "The door cannot be opened." << endl;
		else
			cout << "Ordering is possible." << endl;
	}
	return 0;
}

int find(int x) 
{
	if(x == dp[x])
		return x;
	else
		return find(dp[x]);
}

void dfs(int x,int y) 
{
	int dx = find(x);
	int dy = find(y);
	if(dx != dy)
		dp[dy] = dx;
}

//name:MuaCherish