hdu 6170 Two strings dp

最新推荐文章于 2025-09-12 18:08:49 发布

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最新推荐文章于 2025-09-12 18:08:49 发布

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原文链接：http://www.cnblogs.com/jhz033/p/7412526.html

本文介绍了一个关于两个字符串的匹配问题，给出了具体的输入输出样例，并提供了详细的代码实现。该问题涉及正则表达式的匹配规则，包括特殊符号 . 和 * 的使用。

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;
}