Play on Words - UVa 10129 欧拉回路

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 Specification

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 Specification

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

ac

mmalform

mouse

2

ok

ok

Output for the Sample Input

The door cannot be opened.

Ordering is possible.

The door cannot be opened.

题意：问是否存在一种方式，使得让一个单词的结尾是另一个单词的开头。

思路：就是判断是否有欧拉回路，让字母作为点，除了开头和结尾外，其他点的出度和入度都应该相同，开头和结尾差1（或开头和结尾也是相同字母）。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
char s[1010];
int G[30][30],p[30],in[30],out[30],vis[30];
int find(int x)
{
    return x==p[x] ? x : p[x]=find(p[x]);
}
void Union(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y)
      return;
    p[x]=y;
}
bool check()
{
    int i,ret=0;
    for(i=1;i<=26;i++)
       if(vis[i] && p[i]==i)
         ret++;
    return ret==1;
}
int main()
{
    int T,t,n,i,j,k,u,v,a,b;;
    bool flag;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%d",&n);
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        memset(vis,0,sizeof(vis));
        for(i=1;i<=26;i++)
           p[i]=i;
        for(i=1;i<=n;i++)
        {
            scanf("%s",s+1);
            u=s[1]-'a'+1;
            v=s[strlen(s+1)]-'a'+1;
            out[u]++;
            in[v]++;
            vis[u]=vis[v]=1;
            Union(u,v);
        }
        flag=1;
        if(!check())
          flag=0;
        a=b=0;
        for(i=1;i<=26;i++)
           if(vis[i])
           {
               if(in[i]-out[i]==1)
                  a++;
               else if(out[i]-in[i]==1)
                  b++;
               else if(in[i]!=out[i])
                  flag=0;
           }
        if(a+b>2)
          flag=0;
        if(flag)
          printf("Ordering is possible.\n");
        else
          printf("The door cannot be opened.\n");
    }
}