hdu 6170 Two strings(DP)

Two strings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 855    Accepted Submission(s): 338

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

3aaa*abba.*abbaab

 

Sample Output

yesyesno

 

解题报告：http://blog.csdn.net/jaihk662/article/details/77483118

#include <iostream>#include <cstdio>#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 2500+10;const LL mod = 1e9+7;int dp[N][N];char s1[N], s2[N];int main(){    int t, ncase=1;    scanf("%d", &t);    while(t--)    {        memset(dp,0,sizeof(dp));        scanf("%s %s",s1+1, s2+1);        int l1=strlen(s1+1), l2=strlen(s2+1);        dp[0][0]=1;        for(int i=1;i<=l2;i++)        {            if(i==2&&s2[i]=='*') dp[2][0]=1;//可能*号消除了前面的字符            for(int j=1;j<=l1;j++)            {                if(s2[i]=='.') dp[i][j]=dp[i-1][j-1];                else if(s2[i]!='*')                {                    if(s2[i]==s1[j]) dp[i][j]=dp[i-1][j-1];                }                else                {                    dp[i][j]=max(dp[i-2][j],dp[i-1][j]);                    if(s1[j]==s1[j-1]&&(dp[i-1][j-1]||dp[i][j-1])) dp[i][j]=max(dp[i][j],max(dp[i-1][j-1],dp[i][j-1]));                }            }        }        if(dp[l2][l1]==1) puts("yes");        else puts("no");    }    return 0;}