hdu--（1247）Hat’s Words（trie树）

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8343    Accepted Submission(s): 3004

Problem 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

 

 

Author

戴帽子的

 

简述题意： 给你很多单词，查找能够由其中的两个单词拼合而成的单词，并按字典序输出.....

 

代码：

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<cstdlib>

 4 typedef struct node

 5 {

 6   struct node *child[26];

 7   bool tail;

 8 }Trie;

 9 char mat[50005][15];

10 void Insert(char *s,Trie *root)

11 {

12     int i,pos,j;

13     Trie *cur=root,*curnew;

14     for(i=0;s[i]!='\0';i++){

15         pos=s[i]-'a';

16          if(cur->child[pos]==NULL)

17          {

18              curnew=(Trie *)malloc(sizeof(Trie));

19              for(j=0;j<26;j++)

20              curnew->child[j]=NULL;

21              curnew->tail=false;

22              cur->child[pos]=curnew;

23          }

24          cur=cur->child[pos];

25     }

26      cur->tail=true;

27 }

28 bool check(char *s,Trie *root)

29 {

30     int i,pos;

31     Trie *cur=root;

32    for(i=0;s[i]!='\0';i++)

33    {

34      pos=s[i]-'a';

35      if(cur->child[pos]==NULL) return 0;

36       cur=cur->child[pos];

37    }

38    if(cur->tail==1)return 1;

39    return 0;

40 }

41 bool query(char *s,Trie *root)

42 {

43     int i,pos;

44     Trie *cur=root;

45     for(i=0;s[i]!='\0';i++)

46     {

47         pos=s[i]-'a';

48         if(cur->child[pos]==NULL) return 0;

49         else

50          if(cur->tail==1&&check(s+i,root))

51            return 1;

52          cur=cur->child[pos];

53     }

54     return 0;

55 }

56 int main()

57 {

58    int i=0,j;

59    #ifdef LOCAL

60    freopen("test.in","r",stdin);

61    #endif

62    Trie *root=(Trie *)malloc(sizeof(Trie));

63       root->tail=0;

64      for(j=0;j<26;j++)

65        root->child[j]=NULL;

66    while(scanf("%s",mat[i])!=EOF){

67      Insert(mat[i],root);

68      i++;

69    }

70   for(j=0;j<i;j++)

71   {

72       if(query(mat[j],root))

73         printf("%s\n",mat[j]);

74   }

75   return 0;

76 }

View Code