hdu 1247

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
 
题解：刚开始就老想着在  find函数里修改，其实  模板不用变，就是多了对字符串的处理。
 
代码：
  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <math.h>
  4 #include <algorithm>
  5 #include <iostream>
  6 #include <ctype.h>
  7 #include <iomanip>
  8 #include <queue>
  9 #include <stdlib.h>
 10 using namespace std;
 11 
 12 struct Tri
 13 {
 14     bool v;
 15     Tri* child[26];
 16 };
 17 
 18 Tri* root;
 19 
 20 void Init()
 21 {
 22     root->v=false;
 23     for(int i=0;i<26;i++)
 24     {
 25         root->child[i]=NULL;
 26     }
 27 }
 28 
 29 void CreateDic(char* s)
 30 {
 31     Tri* p;
 32     int j;
 33     int len=strlen(s);
 34     if(len==0)
 35         return ;
 36     p=root;
 37     for(int i=0 ;i < len; i++)
 38     {
 39         if(p->child[s[i]-'a']==NULL)
 40         {
 41             p->child[s[i]-'a']=(Tri*)new Tri;
 42             p->child[s[i]-'a']->v=false;
 43             for(j=0;j<26;j++)
 44                 p->child[s[i]-'a']->child[j]=NULL;
 45         }
 46         p=p->child[s[i]-'a'];
 47         
 48     }
 49     p->v=true;
 50 }
 51 
 52 bool Find(char *s)
 53 {
 54     Tri* p=root;
 55     int len=strlen(s);
 56     if(len==0)
 57         return 0;
 58     for(int i=0 ;i < len; i++)
 59     {
 60         if(p->child[s[i]-'a']==NULL)
 61             return 0;
 62         p=p->child[s[i]-'a'];
 63     }
 64     return p->v;
 65 }
 66 
 67 void Del(Tri* p)
 68 {
 69     for(int i=0;i<26;i++)
 70         if(p->child[i])
 71             Del(p->child[i]);
 72     Del(p);
 73     
 74 }
 75 
 76 char total[50002][100];
 77 int main()
 78 {
 79     char a[100],b[100],c[100];
 80     int i=0,j,k;
 81 
 82     root=(Tri*)new Tri;
 83     Init();
 84     while(gets(a))
 85     {
 86         CreateDic(a);
 87         strcpy(total[i++],a);
 88     }
 89 
 90     for(j=0;j<i;j++)
 91     {
 92         for(k=1;k<strlen(total[j]);k++)
 93         {
 94             strncpy(b,total[j],k);
 95             b[k]='\0';
 96             strcpy(c,total[j]+k);
 97                 
 98             if(Find(b)&&Find(c))
 99             {
100                 cout<<total[j]<<endl;
101                 break;   //注意结束循环                   
102             }
103         }
104     }
105     return 0;
106 }

 

posted @ 2015-12-01 21:06  Vmetrio  阅读(212)  评论(0)    收藏  举报