原題鏈接：[http://acm.hdu.edu.cn/showproblem.php?pid=1501](http://acm.hdu.edu.cn/showproblem.php?pid=1501)
**一：原題內容**
Problem 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
~~~
**二：分析理解**
參考自 [http://www.acmerblog.com/hdu-1501-Zipper-2058.html](http://www.acmerblog.com/hdu-1501-Zipper-2058.html)? 。
最優子結構分析：如上例，如果A、B可以組成C，那么，C最后一個字母e，必定是 A 或 C 的最后一個字母組成。C去除除最后一位，就變成是否可以求出 A-1和B 或者 A與B-1 與 是否可以構成 C-1。。。
? ? ? 狀態轉移方程：用f[i][j] 表示 表示A前 i 為 和B 前j 位是否可以組成 C的前i+j位：dp[i][j]= (dp[i-1][j]&&(a[i]==c[i+j]))||(dp[i][j-1]&&(b[j]==c[i+j]))
**三：AC代碼**
~~~
#include<iostream>
#include<string.h>
using namespace std;
char ch1[202], ch2[202], ch3[404];
int len1, len2, len3;
int DP[202][202];
int main()
{
int N;
cin >> N;
ch1[0] = '1';
ch2[0] = '1';
ch3[0] = '1';
for (int k = 1; k <= N; k++)
{
cin >> ch1 + 1 >> ch2 + 1 >> ch3 + 1;
memset(DP, 0, sizeof(DP));
len1 = strlen(ch1)-1;
len2 = strlen(ch2)-1;
len3 = strlen(ch3)-1;
//處理邊界
for (int i = 1; i <= len1; i++)
{
if (ch1[i] == ch3[i])
DP[i][0] = 1;
}
for (int i = 1; i <= len2; i++)
{
if (ch2[i] == ch3[i])
DP[0][i] = 1;
}
//DP
for (int i = 1; i <= len1; i++)
{
for (int j = 1; j <= len2; j++)
DP[i][j] = DP[i - 1][j] && (ch1[i] == ch3[i + j]) ||
DP[i][j - 1] && (ch2[j] == ch3[i + j]);
}
cout << "Data set " << k << ": ";
if (DP[len1][len2])
cout << "yes\n";
else
cout << "no\n";
}
}
~~~
?
