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

abcdefadebcfabcdefabecdf

 

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<stdio.h>#include<string.h>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");    }}