trie树练习 Hat’s Words

Hat’s Words

TimeLimit: 2000/1000 MS(Java/Others)    MemoryLimit: 65536/32768 K (Java/Others)
Total Submission(s):2174    AcceptedSubmission(s): 791

Problem Description

A hat’s word is a word in the dictionary that is the concatenationof 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 perline, in alphabetical order. There will be no more than 50,000words.
Only one case.

 

Output

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

 

Sample Input

a 

ahat 

hat 

hatword 

hziee 

word

 

Sample Output

ahat 

hatword

#include<stdio.h>

#include<string.h>

#include<malloc.h>

 struct node

{

intflag;

struct node*next[26];

};

struct node*root;

chars[50001][21];

 struct node *chuangjian()

 {

inti;

struct node*p;

p=(structnode*)malloc(sizeof(struct node));

p->flag=0;

for(i=0;i<26;i++)

p->next[i]=NULL;

returnp;

 }

 void insert(char *s)

 {

inti;

struct node*p;

p=(structnode*)malloc(sizeof(struct node));

p->flag=0;

p=root;

intlen=strlen(s);

for(i=0;i<len;i++)

{

if(p->next[s[i]-'a']==NULL)

{

p->next[s[i]-'a']=chuangjian();

}

p=p->next[s[i]-'a'];

}

p->flag=1;

 }

 int  search(char*s)

 {

inti,n;

n=strlen(s);

struct node*p=root;

for(i=0;i<n;i++)

{

if(p->next[s[i]-'a']==NULL)

return0;

elsep=p->next[s[i]-'a'];

}

returnp->flag;

 }

 int main()

 {

intn,i,k,m,j;

charstr1[21],str2[21];

root =(structnode*)malloc(sizeof(struct node));

    for(i=0;i<26;i++)

  root->next[i]=NULL;

k=1;

while(scanf("%s",s[k])!=EOF)

{

insert(s[k]);

k++;

}

for(i=1;i<k;i++)

{

n=strlen(s[i]);

for(j=1;j<n;j++)

{

for(m=0;m<j;m++)//一点点分来测试

str1[m]=s[i][m];

str1[m]='\0';//一定要加上，不然WA

if(search(str1))

{

intp;

for(p=j;p<n;p++)

str2[p-j]=s[i][p];

str2[p-j]='\0';

if(search(str2))

{

printf("%s\n",s[i]);

break;

}

}

}

}

free(root);

return0;

 }