hdu 1247 Hat’s Words 字典树

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Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7870 Accepted Submission(s): 2851

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

aahathathatwordhzieeword

Sample Output

ahathatword

在输入的字符串中如果一个字符由其余的两个字符组成则称为hat‘s words

输出所有的hat‘s words

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<iostream>using namespace std;struct node{    node *next[26];    int end;};node *head;char str[50500][100];void tree_add(char *str){    node *t,*s=head;    int i,j;    int l=strlen(str);    for(i=0;i<l;i++)    {        int id=str[i]-'a';        if(s->next[id]==NULL)        {            t=new node;            for(j=0;j<26;j++)            {                t->next[j]=NULL;            }            t->end=0;            s->next[id]=t;        }        s=s->next[id];    }    s->end=1;}int tree_search(char *str){    int l=strlen(str);    int i,j,id;    node *s=head;    for(i=0;i<l;i++)    {        id=str[i]-'a';        if(s->next[id]==NULL)            return 0;        s=s->next[id];    }    return s->end;}int main(){    int n=0;    head=new node;    int i,j;    char left[100],right[100];    for(i=0;i<26;i++)    {        head->next[i]=NULL;    }    head->end=0;    while(gets(str[n])!=NULL&&strcmp(str[n],"")!=0)    {        tree_add(str[n++]);    }    for(i=0;i<n;i++)    {        int l=strlen(str[i]);        for(j=1;j<l;j++)        {            strcpy(left,str[i]);            left[j]='\0';            strcpy(right,str[i]+j);            if(tree_search(left)&&tree_search(right))            {                puts(str[i]);                break;            }        }    }    return 0;}

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