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.
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.
Your output should contain all the hat’s words, one
per line, in alphabetical order.
ahat
hatword
思路：把每个单词分为两部分，判断两部分是否都在已经建好的字典树中
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<malloc.h>
#define
MAX 50010
using namespace std;
char st[MAX][105];
struct
tree
{
struct tree *child[26];
int v;
}*root;
void init()
{
int
i;
root=(struct tree*)malloc(sizeof(struct
tree));
for(i=0; i<26; i++)
{
root->child[i]=0;
root->v=0;
}
return
;
}
void build(char *str)
{
int
id,i,j,len;
len=strlen(str);
struct
tree *p=root,*q;
for(i=0; i<len;
i++)
{
id=str[i]-‘a‘;
if(p->child[id]==0)
{
q=(struct tree*)malloc(sizeof(struct
tree));
for(j=0; j<26;
j++)
q->child[j]=0;
p->child[id]=q;
p=p->child[id];
p->v=-1;
}
else
{
p=p->child[id];
}
}
p->v=1;
return ;
}
int find(char *str)
{
int
len=strlen(str);
int i,j,id;
struct
tree *p=root;
for(i=0; i<len;
i++)
{
id=str[i]-‘a‘;
if(p->child[id]==0)
return -1;
p=p->child[id];
}
if(p->v==1)
return
1;
else
return -1;
}
int main()
{
int
cnt,i,j,len;
char a[105],b[105];
init();
cnt=0;
while(scanf("%s",st[cnt])!=EOF)
{
build(st[cnt]);
cnt++;
}
for(i=0; i<cnt;
i++)
{
len=strlen(st[i]);
for(j=0;
j<len; j++)
{
memset(
a,‘\0‘,sizeof(a)
);
memset(
b,‘\0‘,sizeof(b)
);
strncpy(a,st[i],j);
strncpy(b,st[i]+j,len-j);
if(find(a)==1&&find(b)==1)
{
printf("%s\n",st[i]);
break;
}
}
}
return 0;
}
