HDU 1116 Play on Words 【并查集+欧拉图】

最新推荐文章于 2018-08-14 15:59:53 发布

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最新推荐文章于 2018-08-14 15:59:53 发布

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本文探讨了一种独特的门锁谜题，其中包含一系列磁性板，每块板上写有一个单词。任务是将这些板排列成序列，使得每个单词的首字母与前一个单词的尾字母相同。文章详细介绍了输入格式、问题描述以及解决方案，包括使用并查集判断是否能形成闭合序列或链，并通过统计入度来确定是否可能打开门锁。

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Play on Words

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5753 Accepted Submission(s): 1903

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 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.

/*
题解：若字符串可以连成环或者成链，则输出Ordering is possible.
运用并查集判断是否是一个图，统计入度判断是否是链或环
*/

#include<cstdio>
#include<cstring>
#define N 100002
int pre[27],in[27];
int find(int x)
{
	return x==pre[x]?x:pre[x]=find(pre[x]);
}

void join(int x,int y)
{
	int fx=find(x),fy=find(y);
	if(fx!=fy)
	{
		pre[fx]=fy;
	}
}

int main()
{
	int T,n,a,b,len,t1,t2,t3,i,flag,vis[27],ok;
	char str[1002];
	scanf("%d",&T);
	while(T--&&scanf("%d",&n))
	{
		for(i=1; i<=26; i++)
		{
			pre[i]=i;
		}
		memset(in,0,sizeof(in));
		memset(vis,0,sizeof(vis));
		for(i=0; i<n; i++)
		{
			scanf("%s",str);
			len=strlen(str);
			a=str[len-1]-'a'+1;
			b=str[0]-'a'+1;
			in[a]--;
			in[b]++;
			vis[a]=vis[b]=1;
			join(a,b);
		}
		for(i=1,t1=0,t2=0,t3=0,flag=0,ok=0; i<=26; i++)
		{
			if(pre[i]==i&&vis[i]) flag++;
			if(in[i]==1) t1++;
			if(in[i]==-1) t2++;
			if(in[i]==0) t3++;
			if(in[i]!=1&&in[i]!=-1&&in[i]!=0) ok=1;
		}
		if(ok)
		{
			printf("The door cannot be opened.\n");
		}
		else if((t1==1&&t2==1&&flag==1)||(t3==26&&flag==1))
		{
			printf("Ordering is possible.\n");
		}
		else
			printf("The door cannot be opened.\n");
	}
	return 0;
}