[欧拉回路] poj 1386 Play on Words

题目链接：

http://poj.org/problem?id=1386

Play on Words

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9685		Accepted: 3344

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

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题目意思：

给n个单词。假设A单词的最后一个字母和B单词的第一个字母是一样的。则说明A和B可以连在一起。

问全部单词是否连成一串。

解题思路：

推断连通性+欧拉通路

仅仅考虑单词的首字母和末尾字母。把出现的字母当成节点，一个单词的首字母和尾字母连成一条边。

先用并查集推断是否连通。然后推断是否有欧拉通路。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 30

bool vis[Maxn];
int dei[Maxn],deo[Maxn],fa[Maxn],n;
char sa[1100];

int Find(int x)
{
    int temp=x;
    while(fa[x]!=x)
        x=fa[x];
    while(fa[temp]!=x)
    {
        int cur=fa[temp];
        fa[temp]=x;
        temp=cur;
    }
    return x;
}
void Unio(int x,int y)
{
    x=Find(x),y=Find(y);
    if(x!=y)
        fa[x]=y;
}

int main()
{
    //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   int t;

   scanf("%d",&t);
   while(t--)
   {
        scanf("%d",&n);
        memset(vis,false,sizeof(vis));
        memset(dei,0,sizeof(dei));
        memset(deo,0,sizeof(deo));

        for(int i=1;i<=26;i++)
            fa[i]=i;

        while(n--)
        {
            scanf("%s",sa+1);
            int len=strlen(sa+1);
            deo[sa[1]-'a'+1]++;
            dei[sa[len]-'a'+1]++;
            vis[sa[1]-'a'+1]=true;
            vis[sa[len]-'a'+1]=true;
            Unio(sa[1]-'a'+1,sa[len]-'a'+1);
        }

        int la=-1,nui=0,nuo=0;
        bool ans=true;

        for(int i=1;i<=26;i++)
        {
            if(vis[i])
            {
                if(la==-1)
                    la=Find(i);
                else if(la!=Find(i))
                {
                     //printf("->i:%d la:%d
",i,la);
                    //system("pause");
                    ans=false;
                    break;
                }
                if(dei[i]-deo[i]==1)
                    nui++;
                else if(deo[i]-dei[i]==1)
                    nuo++;
                else if(deo[i]!=dei[i])
                {
                    ans=false;
                    break;
                }
                //printf("i:%d la:%d
",i,la);
                    //system("pause");


            }
        }
        if(!ans||nui>=2||nuo>=2||(nui+nuo)==1)
            printf("The door cannot be opened.
");
        else
            printf("Ordering is possible.
");


   }
    return 0;
}