一、内容
You are given a string s. You can build new string p from susing the following operation no more than two times: choose any subsequence si1,si2,…,sikwhere 1≤i1<i2<⋯<ik≤|s|;erase the chosen subsequence from s(scan become empty);concatenate chosen subsequence to the right of the string p(in other words, p=p+si1si2…sik ).Of course, initially the string pis empty.For example, let s=ababcd. At first, let's choose subsequence s1s4s5=abc — we will get s=bad and p=abc. At second, let's choose s1s2=ba — we will get s=d and p=abcba. So we can build abcba from ababcd.Can you build a given string tusing the algorithm above?
Input
The first line contains the single integer T(1≤T≤100) — the number of test cases.Next 2Tlines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1≤|s|≤400) — the initial string.The second line contains string tconsisting of lowercase Latin letters (1≤|t|≤|s|) — the string you'd like to build.It's guaranteed that the total length of strings sdoesn't exceed 400
Output
Print Tanswers — one per test case. Print YES (case insensitive) if it's possible to build tand NO (case insensitive) otherwise.
Input
4
ababcd
abcba
a
b
defi
fed
xyz
x
Output
YES
NO
NO
YES
二、思路
三、代码
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 405;
int t, t1n, t2n, n, m, dp[N][N];
char s[N], p[N], t1[N], t2[N];
bool check() {
memset(dp, -1, sizeof(dp)); //-1代表不合法
dp[1][1] = 1; //初始条件
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= t1n + 1; j++) { //最后应该匹配到t1n+1的位置
if (dp[i][j] != -1) {
if (j <=t1n && s[i] == t1[j]) dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j]);
if (dp[i][j] <= t2n && s[i] == t2[dp[i][j]]) dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + 1);
dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
}
}
}
if (dp[n + 1][t1n + 1] > t2n) return true;
return false;
}
void solve() {
scanf("%s%s", s + 1, p + 1);
n = strlen(s + 1), m = strlen(p + 1);
for (int i = 1; i <= m; i++) {
t1n = t2n = 0;
for (int j = 1; j <= i; j++) t1[++t1n] = p[j];
for (int j = i + 1; j <= m; j++) t2[++t2n] = p[j];
if (check()) {printf("YES\n"); return;}
}
printf("NO\n");
}
int main() {
scanf("%d", &t);
while (t--) solve();
return 0;
}
