【递推】【HDU2585】【hotel】

Hotel

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 277    Accepted Submission(s): 161

Problem Description

Last year summer Max traveled to California for his vacation. He had a great time there: took many photos, visited famous universities, enjoyed beautiful beaches and tasted various delicious foods. It is such a good trip that Max plans to travel there one more time this year. Max is satisfied with the accommodation of the hotel he booked last year but he lost the card of that hotel and can not remember quite clearly what its name is. So Max searched 
in the web for the information of hotels in California ans got piles of choice. Could you help Max pick out those that might be the right hotel?

 

Input

Input may consist of several test data sets. For each data set, it can be format as below: For the first line, there is one string consisting of '*','?'and 'a'-'z'characters.This string represents the hotel name that Max can remember.The '*'and '?'is wildcard characters. '*' matches zero or more lowercase character (s),and '?'matches only one lowercase character.

In the next line there is one integer n(1<=n<=300)representing the number of hotel Max found ,and then n lines follow.Each line contains one string of lowercase character(s),the name of the hotel.
The length of every string doesn't exceed 50.

 

Output

For each test set. just simply one integer in a line telling the number of hotel in the list whose matches the one Max remembered.

 

Sample Input

herbert
2
amazon
herbert

?ert*
2
amazon
herbert

*
2
amazon
anything

herbert?
2
amazon
herber

 

Sample Output

1
0 
2
0

 

很好的一个实际中可能出现的问题 关于通配符的

如果利用普通的字符串处理

*a

1

baa 

这种数据十分难处理 不知道a 匹配前一个还是后一个

所以要用递归的方式

#include<stdio.h>  
#include<iostream>  
#include<string>  
using namespace std;  
int find(string a,string b)  
{  
    int i,j;  
    for(i=0;i<a.length();i++)  
    {  
           if(a[i]=='*')  
           {  
                if(i==a.length()-1)  
                        return true;  
               string c=a.substr(i+1);  
                for(j=i;j<b.length();j++)  
                    if(find(c,b.substr(j)))  return 1;  
           }  
           else  
           {  
               if(i>=b.length()) return 0;  
               if(a[i]=='?')  continue;  
               if(a[i]!=b[i]) return 0;  
           }  
    }  
    return 1;  
}  
int main()  
{  
      int ans;  
      string a,b;  
      while(cin>>a)  
      {  
          int k;  
          ans=0;  
          scanf("%d",&k);  
           while(k--)  
           {  
          
               cin>>b;  
               if(find(a,b)) ans+=1;  
           }  
           printf("%d
",ans);  
      }  
      return 0;  
}