Codeforces 1303 E. Erase Subsequences（DP）
Description：
You are given a string s s s. You can build new string p p p from s s s using the following operation no more than two times:
choose any subsequence s i 1 , s i 2 , … , s i k s_{i_{1}},s_{i_{2}},…,s_{i_{k}} si1,si2,…,sik where 1 ≤ i 1 < i 2 < ⋯ < i k ≤ ∣ s ∣ 1≤i_{1}
erase the chosen subsequence from s (s can become empty);
concatenate chosen subsequence to the right of the string p (in other words, p = p + s i 1 s i 2 … s i k p=p+s_{i_{1}}s_{i_{2}}…s_{i_{k}} p=p+si1si2…sik).
Of course, initially the string p p p is empty.
For example, let s = a b a b c d s=ababcd s=ababcd. At first, let’s choose subsequence s 1 s 4 s 5 = a b c s_{1}s_{4}s_{5}=abc s1s4s5=abc — we will get s = b a d s=bad s=bad and p = a b c p=abc p=abc. At second, let’s choose s 1 s 2 = b a s_{1}s_{2}=ba s1s2=ba — we will get s = d s=d s=d and p = a b c b a p=abcba p=abcba. So we can build a b c b a abcba abcba from a b a b c d ababcd ababcd.
Can you build a given string t using the algorithm above?
Input
The first line contains the single integer T ( 1 ≤ T ≤ 100 ) T (1≤T≤100) T(1≤T≤100) — the number of test cases.
Next 2 T 2T 2T lines contain test cases — two per test case. The first line contains string s s s consisting of lowercase Latin letters ( 1 ≤ ∣ s ∣ ≤ 400 ) (1≤|s|≤400) (1≤∣s∣≤400) — the initial string.
The second line contains string t consisting of lowercase Latin letters ( 1 ≤ ∣ t ∣ ≤ ∣ s ∣ ) (1≤|t|≤|s|) (1≤∣t∣≤∣s∣) — the string you’d like to build.
It’s guaranteed that the total length of strings s s s doesn’t exceed 400 400 400.
Output
Print T T T answers — one per test case. Print Y E S YES YES (case insensitive) if it’s possible to build t and N O NO NO (case insensitive) otherwise.
Example
input
4
ababcd
abcba
a
b
defi
fed
xyz
x
output
YES
NO
NO
YES
题意：
给出两个字符串 s s s 和 p p p，对 s s s 可以操作最多两次，从 s s s 中分离出一个子串，然后加到分离出来的串后面，问最多两次操作能不能把 s s s 变成 p p p。
枚举把 t t t 分成哪两段，然后 D P DP DP
把 t t t 分成两个串， f i j f_{ij} fij 表示在 s s s 中匹配第一个串到 i i i， 匹配第二个串 j j j 的最短长度。
其实匹配这一步可以使用序列自动机。
AC代码：
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