Hat’s Words(字典树)

来源：互联网发布：2017最新seo 编辑：程序博客网时间：2024/05/19 00:07

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

题目意思是给你一定的单词，要你找出能够通过其他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    这个题不用也可以过{//    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'};//新建两个临时数组            strncpy(str1,a[i],j);                //把第i个单词进行拆分，分别放入两个临时数组里面            strncpy(str2,a[i]+j,len-j);            if(CZ(str1)&&CZ(str2))            //通过查找函数进行查找，如果两个返回都是1，代表能够组成            {                printf("%s\n",a[i]);                break;            }        }    }    SF(root);//释放内存（这个题可以不要）    return 0;}

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