hdu 5707 Combine String dphen dp

Combine String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2210    Accepted Submission(s): 624

Problem Description

Given three strings  a,  b and  c, your mission is to check whether  c is the combine string of  a and  b.
A string  c is said to be the combine string of  a and  b if and only if  c can be broken into two subsequences, when you read them as a string, one equals to  a, and the other equals to  b.
For example, ``adebcf'' is a combine string of ``abc'' and ``def''.

 

Input

Input file contains several test cases (no more than 20). Process to the end of file.
Each test case contains three strings  a,  b and  c (the length of each string is between 1 and 2000).

 

Output

For each test case, print ``Yes'', if  c is a combine string of  a and  b, otherwise print ``No''.

 

Sample Input

abc def adebcf abc def abecdf

 

Sample Output

Yes

No

很屌丝的题。。

看到的第一眼暴力while走了一遍。。

贪心思想觉得没毛病啊///可是真的有毛病啊。。。

wa了4发。。

得到一组数据

ab

abc

aabcb

abc

ab

aabcb

两个都是Yes

贪心会有一个No

果断dp

dp【i】【j】

i是s1 的某个

j是s2

if(s1[i]==s[i+j])
                    dp[i+1][j]|=dp[i][j];
                if(s2[j]==s[i+j])
                    dp[i][j+1]|=dp[i][j];

#include
#include
int dp[2003][2003];
char s[2003];
char s1[2003];
char s2[2003];
int main()
{
    while(~scanf("%s %s %s",s1,s2,s))
    {
        memset(dp,0,sizeof(dp));
        int Flag=0;
        int len1=strlen(s1);
        int len2=strlen(s2);
        int len=strlen(s);
        dp[0][0]=1;
        for(int i=0;i<=len1;i++)
        {
            for(int j=0;j<=len2;j++)
            {
                

		if(s1[i]==s[i+j])
                    dp[i+1][j]|=dp[i][j];
                if(s2[j]==s[i+j])
                    dp[i][j+1]|=dp[i][j];

} } if(dp[len1][len2]&&len==len1+len2) printf("Yes\n"); else printf("No\n"); }}