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  • 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]='';
     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 }
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  • 原文地址:https://www.cnblogs.com/wangmengmeng/p/5011313.html
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