poj 2337 Catenyms(欧拉路径)

Catenyms

 

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
***

  1 #include<cstdio>
  2 #include<vector>
  3 #include<cstring>
  4 #include<stack>
  5 #include<cstdlib>
  6 #include<algorithm>
  7 using namespace std;
  8 
  9 struct str
 10 {
 11     char word[35];
 12 };
 13 
 14 struct Edge
 15 {
 16     int u,v;
 17 };
 18 
 19 str s[1005];
 20 int in[27],out[27];
 21 bool vis[1005];
 22 vector<Edge>edges;
 23 vector<int>G[1005];
 24 stack<int>S;
 25 int n,beg;
 26 bool flag;
 27 
 28 bool cmp(str a,str b)
 29 {
 30     return strcmp(a.word,b.word)<0;
 31 }
 32 
 33 void init()
 34 {
 35     edges.clear();
 36     while(!S.empty())S.pop();
 37     for(int i=0;i<26;i++)G[i].clear();
 38     memset(in,0,sizeof(in));
 39     memset(out,0,sizeof(out));
 40     memset(vis,0,sizeof(vis));
 41     flag=0;
 42 }
 43 
 44 void dfs(int u)
 45 {
 46     for(int i=0;i<G[u].size();i++)
 47     {
 48         int x=G[u][i];
 49         if(!vis[x])
 50         {
 51             vis[x]=1;
 52             dfs(edges[x].v);
 53             S.push(x);
 54         }
 55     }
 56 }
 57 
 58 void solve()
 59 {
 60     int cnt1=0,cnt2=0;
 61     for(int i=0;i<26;i++)
 62     {
 63         if(in[i]==out[i])
 64             continue;
 65         else if(out[i]-in[i]==1)
 66         {
 67             cnt1++;
 68                 beg=i;
 69             if(cnt1>=2)
 70                 return;
 71         }
 72         else if(in[i]-out[i]==1)
 73         {
 74             cnt2++;
 75             if(cnt2>=2)
 76                 return;
 77         }
 78         else
 79             return;
 80     }
 81     if(cnt1>1||cnt2>1)return;
 82     if(cnt1==0)
 83     for(int i=0;i<26;i++)
 84         if(out[i])
 85         {
 86             beg=i;
 87             break;
 88         }
 89     flag=1;
 90 }
 91 
 92 int main()
 93 {
 94     //freopen("in.txt","r",stdin);
 95     int T;
 96     scanf("%d",&T);
 97     while(T--)
 98     {
 99         init();
100         scanf("%d",&n);
101         for(int i=0;i<n;i++)
102             scanf("%s",s[i].word);
103         sort(s,s+n,cmp);
104         for(int i=0;i<n;i++)
105         {
106             int u=s[i].word[0]-'a';
107             int v=s[i].word[strlen(s[i].word)-1]-'a';
108             in[v]++;
109             out[u]++;
110             edges.push_back((Edge){u,v});
111             G[u].push_back(i);
112         }
113         solve();
114         if(!flag)
115         {
116             printf("***\n");
117             continue;
118         }
119         dfs(beg);
120         if(S.size()!=n)
121         {
122             printf("***\n");
123             continue;
124         }
125         else
126         {
127             bool key=0;
128             while(!S.empty())
129             {
130                 int x=S.top();
131                 if(key)
132                     printf(".");
133                 printf("%s",s[x].word);
134                 key=1;
135                 S.pop();
136             }
137             printf("\n");
138         }
139     }
140     return 0;
141 }

 

posted on 2015-10-31 17:53  焰  阅读(334)  评论(0)  编辑  收藏  举报

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