hdu 6170 Two strings(dp)

Two strings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1420    Accepted Submission(s): 583

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

 

题意;

看两个字符串能否匹配，*表示前面一个字符可以匹配多次，'.'表示匹配可以任意字符，问两个字符能否匹配

解析:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define MAXN 3000char T[MAXN],S[MAXN];int dp[MAXN][MAXN];int n,m;int main(){int t;scanf("%d",&t);while(t--){scanf("%s",S+1);scanf("%s",T+1);n=strlen(S+1);m=strlen(T+1);memset(dp,0,sizeof(dp));dp[0][0]=1;for(int i=1;i<=m;i++){if(T[i]=='*') dp[i][0]=1;  //例如ab,a*ab的情况，dp[2][0]就用于在匹配T[3]时dp[3][1]=dp[2][0],或ab,a*a*ab,dp[5][1]=dp[4][0],dp[3][1]=dp[2][0]for(int j=1;j<=n;j++)   //若i,j可以匹配，那么dp[i][j]由dp[i-1][j-1]匹配的情况决定，若i,j之前的0都能匹配那么dp[i][j]=1{                           //dp[i][j]表示在i,j之前的字符串都成功匹配if(T[i]=='.') dp[i][j]=dp[i-1][j-1];   else if(T[i]!='*') {if(T[i]==S[j]) dp[i][j]=dp[i-1][j-1];}else{dp[i][j]=max(dp[i-1][j],dp[i-2][j]);   //dp[i-1][j]比较大表示*表示次数为1,dp[i-2][j]大表示*的次数为0if(dp[i][j-1]&&S[j]==S[j-1])   //T[i]=='*',*的次数>1的情况{if(T[i-1]=='.'||S[j]==T[i-1]) dp[i][j]=1;  //如果S[j]==T[i-1]('*'前的字符即重复的字符)}}}}if(dp[m][n]) printf("yes\n");else printf("no\n");}return 0;}