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 “*”.
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).
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 ",dp[len1][len]?"yes":"no"); 30 } 31 return 0; 32 }