POJ 2337 Catenyms （有向图欧拉路径，求字典序最小的解）

Catenyms

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8756		Accepted: 2306

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
***

Source

Waterloo local 2003.01.25

 

 

 

 

把26个小写字母当成点，每个单词就是一条边。

然后就是求欧拉路径。

 

 

为了保证字典序最小，要先排序，加边要按照顺序加。

而且求解的dfs起点要选择下，选择最小的。

 

 

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2014-2-3 13:12:43
  4 File Name     :E:\2014ACM\专题学习\图论\欧拉路\有向图\POJ2337.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 struct Edge
 21 {
 22     int to,next;
 23     int index;
 24     bool flag;
 25 }edge[2010];
 26 int head[30],tot;
 27 void init()
 28 {
 29     tot = 0;
 30     memset(head,-1,sizeof(head));
 31 }
 32 void addedge(int u,int v,int index)
 33 {
 34     edge[tot].to = v;
 35     edge[tot].next = head[u];
 36     edge[tot].index = index;
 37     edge[tot].flag = false;
 38     head[u] = tot++;
 39 }
 40 string str[1010];
 41 int in[30],out[30];
 42 int cnt;
 43 int ans[1010];
 44 void dfs(int u)
 45 {
 46     for(int i = head[u] ;i != -1;i = edge[i].next)
 47         if(!edge[i].flag)
 48         {
 49             edge[i].flag = true;
 50             dfs(edge[i].to);
 51             ans[cnt++] = edge[i].index;
 52         }
 53 }
 54 int main()
 55 {
 56     //freopen("in.txt","r",stdin);
 57     //freopen("out.txt","w",stdout);
 58     int T,n;
 59     scanf("%d",&T);
 60     while(T--)
 61     {
 62         scanf("%d",&n);
 63         for(int i = 0;i < n;i++)
 64             cin>>str[i];
 65         sort(str,str+n);//要输出字典序最小的解，先按照字典序排序
 66         init();
 67         memset(in,0,sizeof(in));
 68         memset(out,0,sizeof(out));
 69         int start = 100;
 70         for(int i = n-1;i >= 0;i--)//字典序大的先加入
 71         {
 72             int u = str[i][0] - 'a';
 73             int v = str[i][str[i].length() - 1] - 'a';
 74             addedge(u,v,i);
 75             out[u]++;
 76             in[v]++;
 77             if(u < start)start = u;
 78             if(v < start)start = v;
 79         }
 80         int cc1 = 0, cc2 = 0;
 81         for(int i = 0;i < 26;i++)
 82         {
 83             if(out[i] - in[i] == 1)
 84             {
 85                 cc1++;
 86                 start = i;//如果有一个出度比入度大1的点，就从这个点出发，否则从最小的点出发
 87             }
 88             else if(out[i] - in[i] == -1)
 89                 cc2++;
 90             else if(out[i] - in[i] != 0)
 91                 cc1 = 3;
 92         }
 93         if(! ( (cc1 == 0 && cc2 == 0) || (cc1 == 1 && cc2 == 1) ))
 94         {
 95             printf("***\n");
 96             continue;
 97         }
 98         cnt = 0;
 99         dfs(start);
100         if(cnt != n)//判断是否连通
101         {
102             printf("***\n");
103             continue;
104         }
105         for(int i = cnt-1; i >= 0;i--)
106         {
107             cout<<str[ans[i]];
108             if(i > 0)printf(".");
109             else printf("\n");
110         }
111     }
112     return 0;
113 }