hdu：1247 Hat’s Words

最新推荐文章于 2019-07-25 13:32:11 发布

lkbsbird

最新推荐文章于 2019-07-25 13:32:11 发布

阅读量183

点赞数

CC 4.0 BY-SA版权

分类专栏： hdu 文章标签： hdu

本文链接：https://blog.youkuaiyun.com/lkbsbird/article/details/51842254

hdu 专栏收录该内容

21 篇文章

订阅专栏

本文介绍了一种算法，用于从字典中找出所有由两个其他字典词汇拼接而成的特殊词汇（hat's words）。通过构建前缀树（Trie）并进行搜索，实现了高效查找这些特殊词汇的功能。

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<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=50009;
char sss[maxn][123];
struct trie
{
    int next[32];
    bool num;
    void init()
    {   num=false;
        memset(next,0,sizeof(next));
    }
}tree[maxn*2];
int index;
void insert(char *s)
{
    int len=strlen(s);
    int p=0;
    for(int i=0;i<len;i++)
    {
        int x=s[i]-'a';
        if(tree[p].next[x]==0)
          tree[p].next[x]=++index;
          p=tree[p].next[x];

    }
    tree[p].num=true;
}
int flag=0,vis=0;;
bool find(char *s,int len)
{
    //int len=strlen(s);
    //printf("%s\n",s);
     int p=0;
     for(int i=0;i<len;i++)
     {
         int x=s[i]-'a';
         if(tree[p].next[x]!=0)
          p=tree[p].next[x];
          else
          return false;


     }
     if(tree[p].num)
     return true;
     return false;
}
int main()
{

    int index=0;
    int v[maxn*4];
    tree[0].init();
    while(gets(sss[index])&&sss[index][0])
    {
        insert(sss[index]);
        v[index]=strlen(sss[index]);

       //printf("%s\n",sss[index]);
       index++;
    }
    for(int i=0;i<index;i++)
    {

        //printf("%s \n",sss[i]);
        if(v[i]>1)
        {
            for(int j=1;j<v[i];j++)
             if(find(sss[i],j)&&find(sss[i]+j,v[i]-j))
                {printf("%s\n",sss[i]);
                 break;
                }
        }
    }
    return 0;
}