hdu 6170 Two strings dp

Posted 2020-10-03 xjhz

Two strings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

类似leetcode第10题；

https://leetcode.com/problems/regular-expression-matching/description/

#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)
const int N=3e3+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e18+7;

char s[N],p[N];
bool dp[2510][2510];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(dp,false,sizeof(dp));
        scanf("%s%s",s+1,p+1);
        dp[0][0]=1;
        int n=strlen(s+1);
        int m=strlen(p+1);
        for(int i=2;i<=m;i++)
        {
            dp[0][i]=p[i-1]&&dp[0][i-2];
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(p[j]==‘*‘)
                {
                    dp[i][j] = dp[i][j-2]|| dp[i][j-1] ||( s[i] == p[j-1] || ( p[j-1] == ‘.‘&& s[i]==s[i-1] ) )  && dp[i-1][j];
                }
                else
                dp[i][j] = (s[i] == p[j] || p[j] == ‘.‘) && dp[i-1][j-1];
            }
        }
        if(dp[n][m])printf("yes\n");
        else printf("no\n");
    }
    return 0;
}