Catenyms
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10186 | Accepted: 2650 |
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 *** 先判断是否存在欧拉路径，然后再按照字典序输出欧拉路径，想写个非递归的深搜，可惜失败了
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<string>
5 #include<algorithm>
6 #include<climits>
7 #define MAXE 1010
8 #define MAXP 28
9 using namespace std;
10 struct Edge
11 {
12 int s,t,next;
13 char str[25];
14 }edge[MAXE];
15 int head[MAXE];
16 int degree[MAXP];
17 int fa[MAXP];
18 int stack[MAXE];
19 bool used[MAXP];
20 bool sign[MAXE];
21 bool cur[MAXE][MAXE];
22 int start;
23 int n;
24 int top;
25 bool cmp(Edge a,Edge b)
26 {
27 return strcmp(a.str,b.str)>0;
28 }
29 void add(int s,int t,int ent)
30 {
31 edge[ent].s=s;
32 edge[ent].t=t;
33 edge[ent].next=head[s];
34 head[s]=ent;
35 }
36 int find(int x)
37 {
38 int temp=x,i;
39 while(fa[x]!=x)
40 x=fa[x];
41 while(fa[temp]!=x)
42 {
43 i=fa[temp];
44 fa[temp]=x;
45 temp=i;
46 }
47 return x;
48 }
49 bool oula()
50 {
51 int temp=0,temp2=0;
52 start=edge[n].s;
53 for(int i=1;i<=26;i++)
54 {
55 if(used[i])
56 {
57 if(fa[i]==i)temp++;
58 if(degree[i])
59 {
60 if(degree[i]>1||degree[i]<-1)return false;
61 if(degree[i]==-1)start=i;
62 temp2++;
63 }
64 }
65 }
66 if(temp!=1)return false;
67 if(temp2&&temp2!=2)return false;
68 return true;
69 }
70 void dfs(int s)
71 {
72 for(int i=head[s];i!=-1;i=edge[i].next)
73 {
74 if(!sign[i])
75 {
76 sign[i]=true;
77 dfs(edge[i].t);
78 stack[++top]=i;
79 }
80 }
81 /*while(1)
82 {
83 if(top==n)break;
84 int temp=head[s];
85 for(int i=head[s];i!=-1;temp=i=edge[i].next)
86 {
87 if(!sign[i]&&!cur[stack[top]][i])break;
88 }
89 if(temp==-1)
90 {
91 cur[stack[top-1]][stack[top]]=true;
92 sign[stack[top]]=false;
93 sign[stack[top-1]]=false;
94 s=edge[stack[--top]].s;
95 }
96 else
97 {
98 stack[++top]=temp;
99 s=edge[temp].t;
100 sign[temp]=true;
101 }
102 }*/
103 }
104 int main()
105 {
106 int cas;
107 scanf("%d",&cas);
108 while(cas--)
109 {
110 memset(head,-1,sizeof(head));
111 memset(sign,false,sizeof(sign));
112 memset(used,false,sizeof(used));
113 memset(cur,false,sizeof(cur));
114 memset(degree,0,sizeof(degree));
115 scanf("%d",&n);
116 for(int i=1;i<=n;i++)
117 scanf("%s",edge[i].str);
118 sort(edge+1,edge+1+n,cmp);
119 for(int i=1;i<=26;i++)
120 fa[i]=i;
121 for(int i=1;i<=n;i++)
122 {
123 int s=edge[i].str[0]-‘a‘+1,t=edge[i].str[strlen(edge[i].str)-1]-‘a‘+1;
124 add(s,t,i);
125 fa[find(t)]=find(s);
126 degree[s]--;
127 degree[t]++;
128 used[s]=true;
129 used[t]=true;
130 }
131 if(!oula())
132 {
133 printf("***\n");
134 }
135 else
136 {
137 top=0;
138 dfs(start);
139 /*for(int i=1;i<=top-1;i++)
140 printf("%s.",edge[stack[i]].str);
141 printf("%s\n",edge[stack[top]].str);*/
142 for(int i=top;i>=2;i--)
143 printf("%s.",edge[stack[i]].str);
144 printf("%s\n",edge[stack[1]].str);
145 }
146 }
147 return 0;
148 }