HDU1247 字典树 Hat’s Words（Tire Tree）

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18399    Accepted Submission(s): 6532

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

Author

戴帽子的

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题意：

给定一些单词，要你从中找到某些单词，而这个单词是由另外两个单词组成的。

其实我们就是利用字符的ascii码来给他对应的索引，就是把每个单词拆成两部分，看看是不是每一部分在字典树中都是一个单词

比如说建树的时候存储apple这个单词，对应如下图

在算法导论中，Trie并不叫字典树，而叫基数树，实际上并不是只与字典树有关，是一个N叉树。字典树的功能是对于很多串进行压缩，压缩的方法是跟据这个字符串的前缀，每个节点表示一个字符，从根节点到叶子节点表示一个字符串。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int MAX = 26;
struct Trie
{
     Trie *next[MAX];
     bool isword;
};
Trie *root = new Trie;
char word[50000][30];
void createTrie(char str[])
{
    int len = strlen(str);
    Trie *p = root,*q = NULL;
    for(int i=0;i<len;i++)
    {
         int id = str[i]-'a';
         if(p->next[id]==NULL)
         {
              q = new Trie;
              for(int j=0;j<MAX;j++)
               q->next[j] = NULL;
               q->isword = false;
              p->next[id] = q;
         }
          if(i==len-1)
          p->next[id]->isword = true;
          p = p->next[id];
    }

}
bool findTrie(char str[])
{
     int len = strlen(str);
     Trie *p = root;
   for(int i=0;i<len;i++)
   {
        int id = str[i]-'a';
        if(p->next[id]==NULL)
        {
              return false;
        }
         p = p->next[id];
   }
   if(p->isword)
   return true;
   else
     return false;
}
void del(Trie *root)
{
   for(int i=0;i<MAX;i++)
   {
        if(root->next[i]!=NULL)
        {
             del(root->next[i]);
        }
   }
   delete root;
}
int main()
{
    int num=0;
    char str1[30],str2[30];
    for(int i=0;i<MAX;i++)
    {
         root->next[i] = NULL;
    }
    root->isword = false;
    while(gets(word[num]))
    {
         createTrie(word[num]);
         num++;
    }
    for(int i=0;i<num;i++)
    {
         int len = strlen(word[i]);
         if(len==1)
          continue;
        for(int j=0;j<len;j++)//从每个单词的各部分拆开
     {
          int k;
          if(j==len-1) continue;
         for(k=0;k<=j;k++)
          {
               str1[k] = word[i][k];
          }
          str1[k]='\0';
          int k2=0;
          for(int l=k;l<len;l++)
          {
               str2[k2++]=word[i][l];
          }
          str2[k2]='\0';
          if(findTrie(str1)&&findTrie(str2))
          {
               cout<<word[i]<<endl;
               break;//居然错在这里了(可能会重复输出)
          }
     }
    }
     del(root);
    return 0;
}