Zipper

                           Zipper

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

 

Description

Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.

For example, consider forming "tcraete" from "cat" and "tree":

String A: cat
String B: tree
String C: tcraete


As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "tree":

String A: cat
String B: tree
String C: catrtee


Finally, notice that it is impossible to form "cttaree" from "cat" and "tree".

 

 

Input

The first line of input contains a single positive integer from 1 through 1000. It represents the number of data sets to follow. The processing for each data set is identical. The data sets appear on the following lines, one data set per line.

For each data set, the line of input consists of three strings, separated by a single space. All strings are composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first two strings. The first two strings will have lengths between 1 and 200 characters, inclusive.

 

 

Output

For each data set, print:

Data set n: yes

if the third string can be formed from the first two, or

Data set n: no

if it cannot. Of course n should be replaced by the data set number. See the sample output below for an example.

 

 

Sample Input

3 cat tree tcraete cat tree catrtee cat tree cttaree

 

 

Sample Output

Data set 1: yes

Data set 2: yes

Data set 3: no

【解法一】

 //如果用数组对比，或一般的DFS剪枝，很容易超时

# include<iostream>
using namespace std;
char s1[300],s2[300],OjectStr[600];
int dp[300][300];
int len1,len2,leap;
void dfs(int g1,int g2)
{
    if(leap)return;
    if(g1>len1||g2>len2)return;
    if(g1==len1&&g2==len2)
    {
            leap=1;
            return;
    }
    if(dp[g1][g2])return;//经典
    dp[g1][g2]=1;//如果该个位置没找过，就标记
    if(s1[g1]==OjectStr[g1+g2])dfs(g1+1,g2);
    if(s2[g2]==OjectStr[g1+g2])dfs(g1,g2+1);
}
int main()
{
    int nCase,i;
    cin>>nCase;
    for(i=1;i<=nCase;i++)
    {
       cin>>s1>>s2>>OjectStr;
       len1=strlen(s1);
       len2=strlen(s2);
       leap=0;
       memset(dp,0,sizeof(dp));
       dfs(0,0);
       if(leap)
       cout<<"Data set "<<i<<": yes"<<endl;
       else 
           cout<<"Data set "<<i<<": no"<<endl;
    }
    return 0;
}

 

【解法二】

 //动归，逻辑很清晰

# include<iostream>
# include<cstring>
using namespace std;
char s1[300],s2[300],Str[600];
int dp[300][300];
int len1,len2;
int main()
{
    int nCase,i,j,k;
    cin>>nCase;
    for(k=1;k<=nCase;k++)
    {
       cin>>s1>>s2>>Str;
       len1=strlen(s1);
       len2=strlen(s2);
       memset(dp,0,sizeof(dp));
       dp[0][0]=1;
       for(i=0;i<=len1;i++)
       {
           for(j=0;j<=len2;j++)
           {
               if(i!=0&&dp[i-1][j]&&s1[i-1]==Str[i+j-1])
               {
                   dp[i][j]=1;
               }
               else if(j!=0&&dp[i][j-1]&&s2[j-1]==Str[i+j-1])
               {
                   dp[i][j]=1;
               }
           }
       }
       if(dp[len1][len2])
       cout<<"Data set "<<k<<": yes"<<endl;
       else 
           cout<<"Data set "<<k<<": no"<<endl;
    }
    return 0;
}