OpenJudge Mooc 5.1 Zipper

总时间限制: 

1000ms 

内存限制: 

65536kB

描述

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".

输入

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.

输出

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.

样例输入

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

样例输出

Data set 1: yes
Data set 2: yes
Data set 3: no

/* 
 *  1. dp[i][j]表示 s1的前i位和s2的前j位  组成满足题意的 s3[0..i+j]
    2. 边界条件 dp[0][0] = 1
	3. 状态转移方程：
	   1. if(dp[j - 1][k]==1 && s1[j-1] == s3[j + k -1]) 
	      dp[j][k] = 1;

	   2. if(dp[j][k - 1]==1 && s2[k-1] == s3[j + k-1])
	      dp[j][k] = 1;
*/
#include "iostream"
#include "cstring"
using namespace std;
int main()
{
	char s1[201];
	char s2[201];
	char s3[401];
	int dp[201][201];
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> s1 >> s2 >> s3;
		memset(dp, 0, sizeof(dp));
		dp[0][0] = 1;
		int len1 = strlen(s1);
		int len2 = strlen(s2);
		for (int j = 0; j <= len1; j++)
			for (int k = 0; k <= len2; k++) {
				if (j>0 && dp[j - 1][k]==1 && s1[j-1] == s3[j + k -1])
					dp[j][k] = 1;
				if (k>0 && dp[j][k - 1]==1 && s2[k-1] == s3[j + k-1])
					dp[j][k] = 1;
			}
		printf("Data set %d: ", i);
		if (dp[len1][len2])
			cout << "yes" << endl;
		else
			cout << "no" << endl;
	}
	return 0;
}