Hat’s Words-白红宇

Hat’s Words

阅读量：4971 次

发布时间：2019-06-12

本文共 1500 字，大约阅读时间需要 5 分钟。

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 
        #include<stdio.h> 
#include<malloc.h> 
#include<string.h> 
char lx[50000][50]; 
struct node 
{ 
  bool flag; 
  struct node*next[26]; 
}; 
node root; 
void create(char *s) 
{ 
 node *p=&root,*q; 
 int i,j; 
 int len=strlen(s); 
 for(i=0;i<len;i++) 
 { 
       int num=s[i]-'a'; 
    if(p->next[num]==NULL) 
    { 
     q=(node*)malloc(sizeof(node)); 
     q->flag=false; 
     for(j=0;j<26;j++) 
     { 
      q->next[j]=NULL; 
     } 
     p->next[num]=q; 
    } 
    p=p->next[num]; 
 } 
 p->flag=true; 
} 
bool finder(char *s) 
{ 
   int i,j; 
   for(i=0;i<strlen(s);i++) 
   { 
    node *p=&root; 
       for(j=0;j<i;j++) 
    { 
     int num=s[j]-'a'; 
           p=p->next[num]; 
    } 
    if(p->flag==true) 
    { 
     node *q=&root; 
     bool is=true; 
          for(j=i;j<strlen(s);j++) 
    { 
             int k=s[j]-'a'; 
    if(q->next[k]==NULL) { is=false;break;} 
    q=q->next[k]; 
    } 
    if(q->flag==true&&is==true)  return true; 
    } 
   } 
   return false; 
} 
int main() 
{ 
// freopen("in.txt","r",stdin); 
 int i=0,j; 
 while(scanf("%s",lx[i])!=EOF) 
 { 
  create(lx[i]); 
  i++; 
 } 
 for(j=0;j<i;j++) 
 { 
  if(finder(lx[j]))  printf("%s\n",lx[j]); 
 } 
 return 0; 
} 