HDU 2585 Hotel

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

 

Source

ECJTU 2009 Spring Contest

Solution

給出一個字符串，問下列字符串中與所給匹配的有幾個，？匹配一個字母，*匹配0或多個字母。

一開始想使用暴力掃一遍就好啦，結果*號有些難處理。

模糊匹配的話是可以用簡單dp來做的，f[i][j]代表前i個字符和前j個字符匹配，注意*的話是和後面的所有字符都匹配的

dp的key

f[0][0] = true;//初始化
  for (i = 1; i <= ls; ++i)
    for (j = 1; j <= lx; ++j)
    {
      if ((s[i-1] == x[j-1] || s[i-1] == '?') && f[i-1][j-1]) f[i][j] = true;
      if (s[i-1] == '*' && f[i-1][j-1])
      {
        for (k = j-1; k <= lx; ++k) f[i][k] = true;//可以匹配後面的所有，注意匹配0個的時候是j-1
        break;
      }
    }

後來參考他人的思路，可以用遞歸，這樣比較好理解。

#include <cstdio>
#include <cstring>

char s[55], t[55];
int ls, lt;

bool check(int lens, int lent)
{
  if (lens == ls && lent == lt) return true;
  if (lens == ls || lent == lt) return false;
  if (s[lens] == '?' || s[lens] == t[lent]) return check(lens+1, lent+1);
  if (s[lens] == '*') for (int i = lent; i <= lt; ++i) if (check(lens+1, i)) return true;//往後看看是不是可以匹配
  return false;
}

int main()
{
  while (scanf("%s", s) != EOF)
  {
    ls = strlen(s);
    int n, c = 0;

    scanf("%d", &n);
    while (n--)
    {
      scanf("%s", t);
      lt = strlen(t);
      if (check(0, 0)) ++c;
    }
    printf("%d\n", c);
  }

  return 0;
}