Hat’s Words(字典树)

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

 

题目意思是   给你一定的单词。要你找出可以通过其它两个单词组成的单词。数据有50000多。字典树了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef struct Trie {
    int flag;
    Trie* next[26];
} Trie;
Trie* root;
int flag;
char a[50000][50];
void Inti()//初始化
{
    root = (Trie*)malloc(sizeof(Trie));
    root->flag = 0;
    for (int i = 0; i < 26; i++)
        root->next[i] = NULL;
}
void ChaRu(char *a)//建立字典树    插入单词
{
    Trie* p = root, *q;
    int l = strlen(a);
    for (int i = 0; i < l; i++) {
        int id = a[i] - 'a';
        if (p->next[id] == NULL) {
            q = (Trie *)malloc(sizeof(Trie));
            q ->flag = 0;
            for (int i = 0; i < 26; i++)
                q->next[i] = NULL;
            p->next[id] = q;
        }
       /* else if (p->next[id]->flag) {
            flag = 1;
        }*/
        p = p->next[id];
    }
    p->flag = 1;
}
int CZ(char *a)//查找单词
{
    Trie *p=root;
    int len=strlen(a);
    for(int i=0;i<len;i++)
    {
        int id=a[i]-'a';
        p=p->next[id];
        if(p==NULL)
            return 0;
    }
    return p->flag;
}
int SF(Trie *p)//释放空间    这个题不用也能够过
{
//    Trie *p=root;
    if(p==NULL)return 0;
    for(int i=0;i<26;i++)
    {
        if(p->next[i]!=NULL)
            SF(p->next[i]);
    }
    free(p);
    return 0;
}
int main()
{

    Inti();
    int n=0;
    while (~scanf("%s", &a[n])) {
       // flag = 0;
       //printf("%s",a[i]);
        ChaRu(a[n]);
        n++;
    }

    for(int i=0;i<n;i++)
    {
        int len=strlen(a[i]);
        for(int j=0;j<len;j++)
        {
            char str1[50]={'\0'},str2[50]={'\0'};//新建两个暂时数组
            strncpy(str1,a[i],j);                //把第i个单词进行拆分，分别放入两个暂时数组里面
            strncpy(str2,a[i]+j,len-j);
            if(CZ(str1)&&CZ(str2))            //通过查找函数进行查找。假设两个返回都是1，代表能够组成
            {
                printf("%s\n",a[i]);
                break;
            }
        }
    }
    SF(root);//释放内存（这个题能够不要）
    return 0;
}