HDU 6170 字符串Dp

Two strings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1222    Accepted Submission(s): 507

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 ： 字符串匹配问题 ' . ' 可以匹配任意字符， ‘ * ’ 可以匹配空，或者前面的字符，且可以匹配多个

比如 a * 可以匹配 aaa 和 aaaaaaa等等

这个题贪心是贪不过的，用DP可以，而且字符串长 2500 ，n方的复杂度是允许的

dp[i][j]表示b[i]

对于 ' . ' ，它可以匹配任意字符，所以DP[i[[j] = DP[i - 1][j - 1]

对于' * ' ，它匹配一个的时候状态来自 DP[i - 1][j]  匹配 0 个的时候来自 DP[i - 2][j]

并且 

if((dp[i - 1][j - 1] || dp[i][j - 1]) && a[j - 1] == a[j]){
dp[i][j] = 1;
}

#include<cstdio>#include<algorithm>#include<iostream>#include<string.h>using namespace std;#define maxn 2600int dp[maxn][maxn];char a[maxn],b[maxn];int n;int main(){scanf("%d",&n);getchar();while(n--){memset(dp,0,sizeof(dp));gets(a + 1);gets(b + 1);int lena = strlen(a + 1);int lenb = strlen(b + 1);dp[0][0] = 1;for(int i = 1;i <= lenb;i++){if(i == 2 && b[i] == '*')dp[i][0] = 1;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 - 1][j] | dp[i - 2][j];if((dp[i - 1][j - 1] || dp[i][j - 1]) && a[j - 1] == a[j]){dp[i][j] = 1;}}}}puts(dp[lenb][lena]?"yes":"no");}return 0;}