cf Educational Codeforces Round 81 C. Obtain The String

原题：

C. Obtain The String
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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.
Example
Input

3
aabce
ace
abacaba
aax
ty
yyt

Output

1
-1
3

中文：
给你两个字符串s和t，现在让你每次取s的子串，连接到z串上，使得z与t相同，z初始为空。其中s的子串为在s中删除0个或多个字符串后得到的字符串，问你最少添加多少次。

代码：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 10000 + 1;
 
vector<int> mark[26];
int st[26];
string s, t;
int n;
 
int main()
{
     
	ios::sync_with_stdio(false);
	
	cin >> n;
	while (n--) {
     
		cin >> s >> t;
 
		for (int i = 0; i<26; i++)
			mark[i].clear();
		for (int i = 0; i<s.size(); i++) {
     
			mark[s[i] - 'a'].push_back(i);
		}
		int ans = 1;
		memset(st, 0, sizeof(st));
		int pre; 
		for (int i = 0; i < t.size();) {
     
			if (mark[t[i] - 'a'].empty()) {
     
				ans = -1;
				break;
			}
			if (i == 0) {
     
				pre = mark[t[i] - 'a'].front();// 保存当前下标
				st[t[i]-'a']++;
				i++;
			}
			else {
     
				int flag = 0;
				if (st[t[i] - 'a'] >= mark[t[i] - 'a'].size()) {
     
					flag = 1;
				}
 
				if (!flag) {
     
					// 找大于pre的第一个数的下标
					
					int ind = lower_bound(mark[t[i] - 'a'].begin() + st[t[i] - 'a'], mark[t[i] - 'a'].end(), pre + 1) - mark[t[i] - 'a'].begin();
					if (ind == mark[t[i] - 'a'].size()) {
     
						flag = 1;
					}
					if (!flag) {
     
						pre = mark[t[i] - 'a'][ind];
						st[t[i] - 'a'] = ind + 1;
						i++;
					}
				}
				if(flag) {
     
					pre = -1;
					ans++;
					memset(st, 0, sizeof(st));
				}
				
			}
		}
		
		cout << ans << endl;
 
	}
	
	return 0;
}
 
/*
2
abab
baabaa
abacabadabacaba
baabaab
*/
 
 

解答：

首先把字符串s每个字符对应的下标存储在一个哈希里面，如程序中的mark

遍历字符串t[i]，每次在mark中寻找t[i]在s中对应的下标，因为要找s中的子串，t[i]寻找字符串s中对应的下标一定要大于t[i-1]字符在s中对应的下标，且t[i]在mark中对应的下标要与t[i-1]在mark中对应的下标尽量的靠近（贪心），这里可以使用二分的方式每次寻找字符对应的下标即可。