poj 2192

/**
	2192	Accepted	1244K	63MS	C++	1005B
用布尔变量dp[i][j]表示组成一个字符串用了第一个字符串的i个字符,
用了第二个字符串的j个字符,那么题目所要求的就是dp[len1][len2].
状态转移方程为:dp[i][j]=(((result[i+j-1]==s1[i-1])&&dp[i-1][j])||((result[i+j-1]==s2[j-1])&&dp[i][j-1]))
**/
#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
using namespace std;
string str1,str2,str3;
bool dp[1010][1010];
bool judge()
{
    dp[0][0] = true;
    for(int i = 1; i <= str1.size() ; i++)
    {
        dp[i][0] = ( (str3[i-1] == str1[i-1]) && dp[i-1][0]);
    }
    for(int i = 1; i <= str2.size() ; i++)
    {
        dp[0][i] = ( (str3[i-1] == str2[i-1]) && dp[0][i-1]);
    }
    for(int i = 1; i <= str1.size() ; i++)
        for(int j = 1; j <= str2.size() ;  j++)
        {
            dp[i][j] = ( ( (str3[i+j-1] == str1[i-1]) && dp[i-1][j]) ||((str3[i+j-1] == str2[j-1]) && dp[i][j-1]) );
        }
    return dp[str1.size()][str2.size()];
}
int main()
{
    int n;
    memset(dp,0,sizeof(dp));
    cin >> n;
    for(int nn = 1; nn <= n; nn++)
    {
        cin >> str1 >> str2 >>str3;
        if(judge())
            cout << "Data set "<<nn <<": "<<"yes"<<endl;
        else
            cout << "Data set "<<nn <<": "<<"no"<<endl;


    }
    return 0;
}