trie树练习 Hat’s Words

Hat’s Words

Time
Limit: 2000/1000 MS
(Java/Others)    Memory
Limit: 65536/32768 K (Java/Others)Total Submission(s):
2174    Accepted
Submission(s): 791
Problem DescriptionA 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. InputStandard 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. OutputYour 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#include<stdio.h>
#include<string.h>
#include<malloc.h>
 struct node
{
int
flag;
struct node
*next[26];
};
struct node
*root;
char
s[50001][21];
 struct node *chuangjian()
 {
int
i;
struct node
*p;
p=(struct
node*)malloc(sizeof(struct node));
p->flag=0;
for(i=0;i<26;i++)
p->next[i]=NULL;
return
p;
 }
 void insert(char *s)
 {
int
i;
struct node
*p;
p=(struct
node*)malloc(sizeof(struct node));
p->flag=0;
p=root;
int
len=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)
 {
int
i,n;
n=strlen(s);
struct node
*p=root;
for(i=0;i<n;i++)
{
if(p->next[s[i]-'a']==NULL)
return
0;
else
p=p->next[s[i]-'a'];
}
return
p->flag;
 }
 int main()
 {
int
n,i,k,m,j;
char
str1[21],str2[21];
root =(struct
node*)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))
{
int
p;
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);
return
0;
 }