hdu 6170 Two strings dp模拟

最新推荐文章于 2018-08-23 20:22:45 发布

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最新推荐文章于 2018-08-23 20:22:45 发布

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分类专栏： DP 思路文章标签： dp

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本文介绍了一种解决含通配符的字符串匹配问题的算法，通过模拟递推过程实现两个字符串的有效匹配判断。该算法考虑了特殊符号如'.'和'*'的使用情况，并提供了一个完整的C++代码示例。

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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

题意：给你两个含通配符的字符串，是否能匹配。
注意 a* 可以匹配空字符串
思路：模拟递推过程。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 3005;
char s1[MAXN], s2[MAXN];
int dp[MAXN][MAXN];
int main()
{
    int t;
    scanf("%d", &t);
    getchar();
    while(t--)
    {
        gets(s1+1);
        gets(s2+1);
        int len1=strlen(s1+1);
        int len2=strlen(s2+1);
        memset(dp, 0, sizeof(dp));
        dp[0][0]=true;
        for(int i=1; i<=len2; ++i)
        {
            for(int j=1; j<=len1; ++j)
            {
                if(s2[i]==s1[j]||s2[i]=='.')
                    dp[i][j]|=dp[i-1][j-1];
                else if(s2[i]=='*')
                {
                    if(s1[j]==s1[j-1])
                        dp[i][j]|=dp[i][j-1];//可以是代表多个字符
                    dp[i][j]|=dp[i-1][j];//可以是一个字符，a*为a
                    dp[i][j-1]|=dp[i-2][j-1];//可以是空字符;
                }
            }
        }
        if(dp[len2][len1])
            puts("yes");
        else
            puts("no");
    }
    return 0;
}