HDU 6170 dp
Two strings
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 485 Accepted Submission(s): 178
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 “*”.
For each test case, there are two lines implying the two strings (The length of the two strings is less than 2500).
aa
a*
abb
a.*
abb
aab
yes
no
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
#define PI acos(-1.0)
int t;
char a[],b[];
int dp[][];
int main()
{
scanf("%d",&t);
while(t--){
scanf("%s",a+);
scanf("%s",b+);
memset(dp,,sizeof(dp));
int lena=strlen(a+);
int lenb=strlen(b+);
dp[][]=;
for(int i=;i<=lenb;i++){
if(i==&&b[i]=='*')
dp[i][]=;
for(int j=;j<=lena;j++){
if(b[i]=='.'||b[i]==a[j])
dp[i][j]=dp[i-][j-];
else if(b[i]=='*'){
dp[i][j]=dp[i-][j]|dp[i-][j];
if((dp[i-][j-]||dp[i][j-])&&a[j-]==a[j])
dp[i][j]=;
}
}
}
if(dp[lenb][lena]==)
printf("yes\n");
else
printf("no\n");
}
return ;
}
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