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  • hdu 6170 Two strings dp

    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

    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
    ");
            else printf("no
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/7412526.html
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