HDU 1247 Hat’s Words

Description

A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.
You are to find all the hat’s words in a dictionary.

Input

Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 50,000 words.
Only one case.

Output

Your output should contain all the hat’s words, one per line, in alphabetical order.

Sample Input

a
ahat
hat
hatword
hziee
word

Sample Output

ahat
hatword

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cstdlib>
 4 typedef struct Trie
 5 {
 6     int num;
 7     bool excist;
 8     struct Trie *next[26];
 9 }Node,*trie;
10 char ss[500000][15];
11 Node *create()
12 {
13     Node *node=(Node *)malloc(sizeof(Node));
14     node->num=0;
15     node->excist=false;
16     memset(node->next,0,sizeof(node->next));
17     return node;
18 }
19 void insert_(trie root,char *str)
20 {
21     trie node=root;
22     char *p=str;
23     int id;
24     while(*p)
25     {
26         id=*p-'a';
27         if(node->next[id]==NULL)
28         {
29             node->next[id]=create();
30         }
31         node=node->next[id];
32         ++p;
33         node->num+=1;
34     }
35     node->excist=true;
36 }
37  int finds(trie root,char *str)
38 {
39     trie node=root;
40     char *p=str;
41     int id;
42     while(*p)
43     {
44         id=*p-'a';
45         node=node->next[id];
46         ++p;
47         if(node==NULL)
48             return 0;
49     }
50     return node->excist;
51 }
52 int main()
53 {
54     trie root=create();
55     char str[15];
56     int k=0;
57     while(scanf("%s",str)>0)
58     {
59         //if(strlen(str)==0)
60             //break;
61         strcpy(ss[k++],str);
62         insert_(root,str);
63     }
64     char s1[15],s2[15];
65     for(int i=0;i<k;i++)
66     {
67         int len=strlen(ss[i]);
68         for(int j=1;j<len;j++)
69         {
70             strcpy(s1,ss[i]);
71             //printf("%s %s\n",s1,ss[i]);
72             s1[j]='\0';
73             //printf("%s %s\n",s1,ss[i]);
74             //printf("%s\n",s1[j]);
75             strcpy(s2,ss[i]+j);
76             //printf("%s %s\n",s2,ss[i]+j);
77             if(finds(root,s1)&&finds(root,s2))
78             {
79                 printf("%s\n",ss[i]);
80                 break;
81             }
82         }
83     }
84     return 0;
85 }

 

posted @ 2016-09-23 13:09  PrayG  阅读(146)  评论(0)  编辑  收藏  举报