hdu 1247 Hat’s Words 字典树

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

 

         字典树：以给的每个单词建树。然后给每个单词分成两半查找，如果两半都在树中查找到了，就输出者个单词。

 

     

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[50005][50];
struct f
{
    int next[30];
    int v;
    void ff()
    {
        memset(next,-1,sizeof(next));
        v=0;
    }
};f tree[1000000];
int cut=0;
void buld(char *s1)
{
    int i,k=0,su;
    for (i=0;s1[i]!='';i++)
    {
        su=s1[i]-'a';
        if (tree[k].next[su]==-1)
        {
            cut++;
            tree[k].next[su]=cut;
            tree[cut].ff();
        }
        k=tree[k].next[su];
    }
    tree[k].v=1;
    return ;
}
int find(char *s1)
{
    int i,k=0,su,len;
    len=strlen(s1);
    for (i=0;i<len;i++)
    {
        su=s1[i]-'a';
        if (tree[k].next[su]==-1) break;
        k=tree[k].next[su];
    }
    if (i==len&&tree[k].v) return 1;
    return 0;
}
int main()
{
     int p=0,i,j,len,k,d;
     char sl[50],sr[50];
     tree[cut].ff();
     while (~scanf("%s",s[p]))
     {
         buld(s[p]);
         p++;
     }
     for (i=0;i<p;i++)
     {
         len=strlen(s[i]);
         for (j=0;j<len;j++)
         {
             memset(sr,'',sizeof(sr));
             memset(sl,'',sizeof(sl));
            for (k=0;k<j;k++) sl[k]=s[i][k];
            sl[j]='';
            for (k=j,d=0;k<len;k++,d++) sr[d]=s[i][k];
            sr[len]='';
             if (find(sl)&&find(sr))
             {
                 printf("%s
",s[i]);
                 break;
             }
         }
     }
     return 0;
}