poj2337欧拉回路要求输出路径

                     Catenyms

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8368		Accepted: 2202

Description

A catenym is a pair of words separated by a period such that the last letter of the first word is the same as the last letter of the second. For example, the following are catenyms:

dog.gopher
 gopher.rat
 rat.tiger
 aloha.aloha
 arachnid.dog

A compound catenym is a sequence of three or more words separated by periods such that each adjacent pair of words forms a catenym. For example,
aloha.aloha.arachnid.dog.gopher.rat.tiger
Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once.

Input

The first line of standard input contains t, the number of test cases. Each test case begins with 3 <= n <= 1000 - the number of words in the dictionary. n distinct dictionary words follow; each word is a string of between 1 and 20 lowercase letters on a line by itself.

Output

For each test case, output a line giving the lexicographically least compound catenym that contains each dictionary word exactly once. Output "***" if there is no solution.

Sample Input

2
6
aloha
arachnid
dog
gopher
rat
tiger
3
oak
maple
elm

Sample Output

aloha.arachnid.dog.gopher.rat.tiger
***

用DFS实现的

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <string.h>
 #include <math.h>
 #include <vector>
 #include <stack>
 #include <map>
 using namespace std;
 #define ll long long int
 #define INF 5100000
 string a[];
 int v[];
 vector<pair<int,int> >aa[];
 vector<int>ab;
 int p[];
 int d1[];
 int d2[];
 int find(int x)
 {
     while(x!=p[x])
         x=p[x];
     return x;
 }
 void fun(int x,int eid)
 {
     for(int i=; i<aa[x].size(); i++)
         if(!v[aa[x][i].second])
         {
             v[aa[x][i].second]=true;
             fun(aa[x][i].first,aa[x][i].second);
         }
     if(eid>)ab.push_back(eid);
 }
 int main()
 {
     int tt;
     int i,j;
     cin>>tt;
     for(i=; i<tt; i++)
     {
         int n;
         cin>>n;
         memset(d1,,sizeof(d1));
         memset(d2,,sizeof(d2));
         for(j=; j<; j++)
             p[j]=j,aa[j].clear();
         for(j=; j<=n; j++)
         {
             cin>>a[j];
         }
         sort(a+,a+n+);
         for(j=; j<=n; j++)
         {
             int x=a[j][]-'a'+;
             int y=a[j][a[j].length()-]-'a'+;
             d1[x]++;
             d2[y]++;
             aa[x].push_back(make_pair(y,j));
             int fx=find(x);
             int fy=find(y);
             if(fy!=fx)
                 p[fy]=fx;
         }
         int sum=;
         int start=a[][]-'a'+;
         for(j=; j<; j++)
         {
             if(d1[j]==&&d2[j]==)continue;
             if(p[j]==j)sum++;
         }
         if(sum>)
             cout<<"***"<<endl;
         else
         {
             int dd1=,dd2=;
             for(j=; j<; j++)
             {
                 if(d1[j]!=d2[j])
                 {
                     if(d1[j]-d2[j]==)
                         dd1++,start=j;
                     else if(d2[j]-d1[j]==)
                         dd2++;
                     else
                     {
                         dd2=;
                         break;
                     }
                 }
             }
             if(dd2<=&&dd1<=)
             {
                 memset(v,,sizeof(v));
                 ab.clear();
                 fun(start,);
                 for( j=ab.size()-; j>=; j--)
                 {
                     cout<<a[ab[j]];
                     if(j!=)cout<<".";
                     else cout<<"\n";
                 }

             }
             else cout<<"***"<<endl;
         }
     }
 }