HDU 6170Two strings

最新推荐文章于 2019-10-31 23:24:41 发布

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最新推荐文章于 2019-10-31 23:24:41 发布

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原文链接：http://www.cnblogs.com/xingkongyihao/p/7413929.html

本文介绍了一种使用动态规划解决正则表达式匹配问题的方法。通过构建二维DP数组来判断两个字符串是否匹配，其中涉及特殊字符'.'和'*'的处理逻辑。最终实现了对于包含特定正则表达式的字符串的有效匹配。

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Problem Description

Giving two strings and you should judge if they are matched.
The first string contains lowercase letters and uppercase letters.
The second string contains lowercase letters, uppercase letters, and special symbols: “.” and “*”.
. can match any letter, and * means the front character can appear any times. For example, “a.b” can match “acb” or “abb”, “a*” can match “a”, “aa” and even empty string. ( “*” will not appear in the front of the string, and there will not be two consecutive “*”.

Input

The first line contains an integer T implying the number of test cases. (T≤15)
For each test case, there are two lines implying the two strings (The length of the two strings is less than 2500).

Output

For each test case, print “yes” if the two strings are matched, otherwise print “no”.

Sample Input

3

aa

a*

abb

a.*

abb

aab

Sample Output

yes

yes

no

正则表达式的问题。

题解说的是用动态规矩。

str表示主串，str1表示模拟串，可以假设dp[i][j]表示str1[1,i]和str[1,j]是否匹配。

显然dp[0][0] = true.

str1[i] == . 或者str1[i] == str[j]时，dp[i][j] 看状态dp[i-1][j-1]

str1[i] == '*'时，dp[i][j] == dp[i-1][j] | dp[i-2][j],而当(dp[i-1][j-1] || dp[i][j-1]) && str[j-1] == str[j] 时，dp[i][j]必定为true；

详细参考这

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 using namespace std;
 6 const int N = 2510;
 7 char str[N], str1[N];
 8 bool dp[N][N];
 9 int main() {
10     int t;
11     scanf("%d", &t);
12     while(t--) {
13         memset(dp, false, sizeof(dp));
14         scanf("%s %s",str+1, str1+1);
15         int len = strlen(str+1), len1 = strlen(str1+1);
16         dp[0][0] = true;
17         for(int i = 1; i <= len1; i ++) {
18             if(i == 2 && str1[i] == '*') dp[i][0] = true;
19             for(int j = 1; j <= len; j ++) {
20                 if(str1[i] == '.' || str1[i] == str[j])
21                     dp[i][j] = dp[i-1][j-1];
22                 else if(str1[i] == '*') {
23                     dp[i][j] = dp[i-2][j] | dp[i-1][j];
24                     if((dp[i-1][j-1] || dp[i][j-1]) && str[j-1] == str[j])
25                         dp[i][j] = true;
26                 }
27             }
28         }
29         printf("%s\n",dp[len1][len]?"yes":"no");
30     }
31     return 0;
32 }