HDU 6170 Two strings (dp)

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

 

题意

给出原串与匹配串,问能否匹配原串中所有的字符。

 

思路

如果这是一个标准的正则匹配,是不是可以直接用语言特性了呢?

我们设原串为 a ,匹配串为 bdp[i][j] 代表 b[1..i]a[1..j] 是否匹配成功。

显然 dp[0][0] = true

对于其他情况:

  • 如果 b[i] == '.' ,则此时 a[j] 可以是任意字符, dp[i][j]dp[i-1][j-1] 转移而来。
  • 如果 a[j] == b[i] ,同样 dp[i][j]dp[i-1][j-1] 转移而来。

  • 如果 b[i] == '*' ,假设该 * 最终可以匹配 0 位,则 dp[i][j] 状态从 dp[i-2][j] 转移而来,假设最终匹配 1 位,则从 dp[i-1][j] 转移而来;

    假如 a[1..j-1]b[1..i-1] 已成功匹配,并且 a[j-1] == a[j] ,显然当前的 * 可以继续匹配这一个字符,因此 dp[i][j] = true

    假如 a[1..j-1]b[1..i] 已成功匹配(当前 * 已成功匹配若干位),且 a[j-1] == a[j] ,则可以继续匹配这一个字符,因此 dp[i][j] = true

  • 特别的,如果 b[i] == '*' ,则 dp[i][0] 可以从 dp[i-2][0] 转移而来,因此 dp[i][0] |= dp[i-2][0]

 

AC 代码

#include <bits/stdc++.h>
using namespace std;
const int maxn=2600;

char a[maxn],b[maxn];
bool dp[maxn][maxn];
int lena,lenb;
int main()
{
    int T;
    scanf("%d%*c",&T);
    while(T--)
    {
        memset(dp,false,sizeof(dp));
        a[0]=b[0]=1;
        gets(a+1);
        gets(b+1);
        lena = strlen(a) - 1;
        lenb = strlen(b) - 1;
        dp[0][0]=true;
        for(int i=1; i<=lenb; i++)
        {
            if(i>=2&&b[i]=='*')
                dp[i][0] |= dp[i-2][0];
            for(int j=1; j<=lena; j++)
            {
                if(b[i]=='.'||a[j]==b[i])
                    dp[i][j]=dp[i-1][j-1];
                else if(b[i]=='*')
                {
                    dp[i][j]=dp[i-2][j]|dp[i-1][j];
                    if((dp[i-1][j-1]||dp[i][j-1])&&a[j-1]==a[j])
                        dp[i][j] = true;
                }
            }
        }
        puts(dp[lenb][lena]?"yes":"no");
    }
    return 0;
}