HDU 2585 Hotel（字符串的模糊匹配+递归）

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

 

启发博客：http://blog.csdn.net/light_14/article/details/43940765

字符串的模糊匹配，？可以当做都匹配的一个字符，遇到*时自带一个查找

 

 #include <iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<string>

 using namespace std;

 char a[],b[];
 int n,len1,len2,res;

 bool digital(int i,int j)
 {
     //结尾判定
     if(i==len1&&j==len2)
         return true;
     else if(i==len1||j==len2)
         return false;
     //中间过程
     else if(a[i]=='*')
     {
         for(int k=j;k<=len2;k++)
             if(digital(i+,k))
             return true;
     }
     else if(a[i]=='?'||a[i]==b[j])
         digital(i+,j+);
     else
         return false;
 }

 int main()
 {
     while(scanf("%s",&a)!=EOF)
     {
         scanf("%d\n",&n);
         len1=strlen(a);
         res=;
         while(n--)
         {
             scanf("%s",&b);
             len2=strlen(b);
             if(digital(,))
                 res++;
         }
         printf("%d\n",res);
     }
     return ;
 }