Play on Words[HDU1116]

Play on Words

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4094    Accepted Submission(s): 1328

 

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.
 

 

Source
Central Europe 1999
 

 

Recommend
Eddy

Take letters 'a'..'z' as the nodes in graph.A word start with charcateristic c1 and end by characteristic c2 forms an edge in the graph.Then the problem become if there exist an Euler circuit.
So how can we judge whether a graph exists an Euler circuit or not ? First the graph should be a connected graph.We can determine this by breadth-first search.Then we have to check the in-degree and out-degree.Only two or zero if them can be different.If there are two difference,one of the two its in-degree minus its out-degree should be one while the other should be minus one.

 

#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
using namespace std;
int dr[26],dc[26],N,f[26];
bool e[26],g[26][26];
bool bfs()
{
	int i;
	memset(f,0,sizeof(f));
	queue<int> q;
	while (!q.empty()) q.pop();
	for (i=0;i<26;i++)
		if (e[i])
		{
			f[i]=true;
			q.push(i);
			break;
		}
	while (!q.empty())
	{
		int x=q.front();
		q.pop();
		for (i=0;i<26;i++)
			if (g[x][i])
			{
				if (!f[i])
				{
					f[i]=true;
					q.push(i);
				}
			}
	}
	for (i=0;i<26;i++)
		if ((e[i]) && (!f[i])) return false;
	return true;
}
int main()
{
	int T,i;
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d",&N);
		char s[1024];
		memset(g,0,sizeof(g));
		memset(e,0,sizeof(e));
		memset(dr,0,sizeof(dr));
		memset(dc,0,sizeof(dc));
		for (i=1;i<=N;i++)
		{
			scanf("%s",s);
			int u=s[0]-'a',v=s[strlen(s)-1]-'a';
			g[u][v]=true;
			g[v][u]=true;
			e[u]=true;
			e[v]=true;
			dc[u]++;
			dr[v]++;
		}
		bool flag=bfs();
		int cnt0=0,cnt1=0;
		for (i=0;i<26;i++)
		{
			if (dr[i]-dc[i]>1 || dr[i]-dc[i]<-1) flag=false;
			if (dr[i]-dc[i]==1) cnt0++;
			if (dr[i]-dc[i]==-1) cnt1++;
		}
		if (!((cnt0==0 && cnt1==0)||(cnt0==1 && cnt1==1))) flag=false;
		if (flag) printf("Ordering is possible.
");
		else printf("The door cannot be opened.
");
	}
	return 0;
}