HDU6170-Two strings

Two strings

                                                                    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                              Total Submission(s): 492    Accepted Submission(s): 183

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

3aaa*abba.*abbaab

 

Sample Output

yesyesno

 

Source

2017 Multi-University Training Contest - Team 9

 

题意：有两个字符串，问你第二个字符串和第一个字符串能否匹配，第二个字符串有两种符号，’.’可以匹配任意字符，’*’表示前一个字符可以重复零次或多次

解题思路：比赛时发现了可以n^2跑，可是居然没想到dp，一直想着暴力匹配，太菜了，dp还是很好推的，dp[i][j]表示第二个字符串前i个字符和第一个字符串前j个字符能否匹配

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <map>  #include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  #include <functional>  using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;int dp[2505][2505];char s1[2505], s2[2505];int main(){int t;scanf("%d", &t);while (t--){scanf("%s%s", s1 + 1, s2 + 1);memset(dp, 0, sizeof dp);int len1 = strlen(s1 + 1), len2 = strlen(s2 + 1);dp[0][0] = 1;for (int i = 1; i <= len2; i++){if (s2[i] == '*' && i == 2) dp[i][0] = 1;for (int j = 1; j <= len1; j++){if (s2[i] == '.') dp[i][j] = dp[i - 1][j - 1];else if (s2[i] != '*'){if (s2[i] == s1[j])dp[i][j] = dp[i - 1][j - 1];}else{dp[i][j] = max(dp[i - 2][j], dp[i - 1][j]);if (dp[i][j - 1] && s1[j] == s1[j - 1]){if (s2[i - 1] == s1[j] && s2[i - 1] != '.') dp[i][j] = 1;else if (s2[i - 1] == '.') dp[i][j] = 1;}}}}if (dp[len2][len1]) printf("yes\n");else printf("no\n");}return 0;}