CF 1295——C.Obtain The String【二分】

题目传送门

---

You are given two strings s and t consisting of lowercase Latin letters. Also you have a string z which is initially empty. You want string z to be equal to string t. You can perform the following operation to achieve this: append any subsequence of s at the end of string z. A subsequence is a sequence that can be derived from the given sequence by deleting zero or more elements without changing the order of the remaining elements. For example, if $z=ac, s=abcde$, you may turn z into following strings in one operation:

$z=acace$ (if we choose subsequence ace);
$z=acbcd$ (if we choose subsequence bcd);
$z=acbce$ (if we choose subsequence bce).
Note that after this operation string s doesn’t change.

Calculate the minimum number of such operations to turn string z into string t.

---

Input

The first line contains the integer $T (1≤T≤100)$ — the number of test cases.

The first line of each testcase contains one string $s (1≤|s|≤10^5)$ consisting of lowercase Latin letters.

The second line of each testcase contains one string $t (1≤|t|≤10^5)$ consisting of lowercase Latin letters.

It is guaranteed that the total length of all strings s and t in the input does not exceed $2 *10^5$.

---

Output

For each testcase, print one integer — the minimum number of operations to turn string z into string t. If it’s impossible print −1.

---

input

3
aabce
ace
abacaba
aax
ty
yyt

---

output

1
-1
3

---

题意

• 给定两个字符串s和t，你有一个空字符串z

• 每次可以取s的任意一个子序列加到z后面

• 问至少要取多少次才能让z等价于t

---

题解

• vector存s中26个字母的位置

  然后字符串 $t$从前往后一个个查找

  用变量p记录查到上一个字符时在字符串s中的位置（初始化为-1）

  如果在t内碰到一个字符，没有在s中出现过，输出-1结束当次程序

  否则，看当前的p是否大于等于这个字符在s中出现的最后一个位置

  如果是，说明这一次子序列已经不能往后面取了，说明得另起一次从头取子序列

  ans++，让p等于当前字符在s中出现的第一个位置

  如果不是二分找大于p的第一个这个字符的位置，让p等于这个位置即可

• 注意，ans初始化为1，vector要清空……

  思路用到了一点点贪心，就是每次都是取上一次的位置后出现的第一个指定字符的位置

  反正正常模拟肯定超时要二分优化，嗯！

---

AC-Code

#include <bits/stdc++.h>
using namespace std;

vector<int>vec[26];
int num[26];
int main() {
	int T;
	cin >> T;
	while (T--) {
		memset(num, 0, sizeof num);
		for (int i = 0; i < 26; ++i)
			vec[i].clear();
		string s, t;
		cin >> s >> t;
		for (int i = 0; i < s.length(); ++i) {
			vec[s[i] - 'a'].push_back(i);
			++num[s[i] - 'a'];
		}
		int ans = 1;
		int p = -1;
		for (int i = 0; i < t.length(); ++i) {
			if (!num[t[i] - 'a']) {
				ans = -1;
				break;
			}
			else if (p >= vec[t[i] - 'a'][num[t[i] - 'a'] - 1]) {
				++ans;
				p = vec[t[i] - 'a'][0];
			}
			else 
				p = vec[t[i] - 'a'][upper_bound(vec[t[i] - 'a'].begin(), vec[t[i] - 'a'].end(), p) - vec[t[i] - 'a'].begin()];
		}
		cout << ans << endl;
	}

	return 0;
}