HDU 6170 dp

Posted 2020-10-03 半根毛线code

Two strings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 485    Accepted Submission(s): 178

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

 

 

Source

2017 Multi-University Training Contest - Team 9

 千千详细分析

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<set>
 7 #include<map>
 8 #include<queue>
 9 #include<stack>
10 #include<vector>
11 using namespace std;
12 typedef long long ll;
13 #define PI acos(-1.0)
14 int t;
15 char a[2505],b[2505];
16 int dp[2505][2505];
17 int main()
18 {
19     scanf("%d",&t);
20     while(t--){
21         scanf("%s",a+1);
22         scanf("%s",b+1);
23         memset(dp,0,sizeof(dp));
24         int lena=strlen(a+1);
25         int lenb=strlen(b+1);
26         dp[0][0]=1;
27         for(int i=1;i<=lenb;i++){
28             if(i==2&&b[i]==‘*‘)
29              dp[i][0]=1;
30             for(int j=1;j<=lena;j++){
31                 if(b[i]==‘.‘||b[i]==a[j])
32                     dp[i][j]=dp[i-1][j-1];
33                 else if(b[i]==‘*‘){
34                     dp[i][j]=dp[i-2][j]|dp[i-1][j];
35                     if((dp[i-1][j-1]||dp[i][j-1])&&a[j-1]==a[j])
36                     dp[i][j]=1;
37                 }
38             }
39         }
40         if(dp[lenb][lena]==1)
41             printf("yes\n");
42         else
43             printf("no\n");
44     }
45     return 0;
46 }