A13616. String Transformation 1
编程题
普及/提高-
知识点
题目描述
Note that the only difference between String Transformation 1 and String Transformation 2 is in the move Koa does. In this version the letter $y$ Koa selects must be strictly greater alphabetically than $x$ (read statement for better understanding). You can make hacks in these problems independently.
Koa the Koala has two strings $A$ and $B$ of the same length $n$ ( $|A|=|B|=n$ ) consisting of the first $20$ lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions $p_1, p_2, \ldots, p_k$ ( $k \ge 1; 1 \le p_i \le n; p_i \neq p_j$ if $i \neq j$ ) of $A$ such that $A_{p_1} = A_{p_2} = \ldots = A_{p_k} = x$ (ie. all letters on this positions are equal to some letter $x$ ).
2. selects a letter $y$ (from the first $20$ lowercase letters in English alphabet) such that $y>x$ (ie. letter $y$ is strictly greater alphabetically than $x$ ).
3. sets each letter in positions $p_1, p_2, \ldots, p_k$ to letter $y$ . More formally: for each $i$ ( $1 \le i \le k$ ) Koa sets $A_{p_i} = y$ . Note that you can only modify letters in string $A$ .
Koa wants to know the smallest number of moves she has to do to make strings equal to each other ( $A = B$ ) or to determine that there is no way to make them equal. Help her!
Koa the Koala has two strings $A$ and $B$ of the same length $n$ ( $|A|=|B|=n$ ) consisting of the first $20$ lowercase English alphabet letters (ie. from a to t).
In one move Koa:
1. selects some subset of positions $p_1, p_2, \ldots, p_k$ ( $k \ge 1; 1 \le p_i \le n; p_i \neq p_j$ if $i \neq j$ ) of $A$ such that $A_{p_1} = A_{p_2} = \ldots = A_{p_k} = x$ (ie. all letters on this positions are equal to some letter $x$ ).
2. selects a letter $y$ (from the first $20$ lowercase letters in English alphabet) such that $y>x$ (ie. letter $y$ is strictly greater alphabetically than $x$ ).
3. sets each letter in positions $p_1, p_2, \ldots, p_k$ to letter $y$ . More formally: for each $i$ ( $1 \le i \le k$ ) Koa sets $A_{p_i} = y$ . Note that you can only modify letters in string $A$ .
Koa wants to know the smallest number of moves she has to do to make strings equal to each other ( $A = B$ ) or to determine that there is no way to make them equal. Help her!
输入格式
Each test contains multiple test cases. The first line contains $t$ ( $1 \le t \le 10$ ) — the number of test cases. Description of the test cases follows.
The first line of each test case contains one integer $n$ ( $1 \le n \le 10^5$ ) — the length of strings $A$ and $B$ .
The second line of each test case contains string $A$ ( $|A|=n$ ).
The third line of each test case contains string $B$ ( $|B|=n$ ).
Both strings consists of the first $20$ lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$ .
The first line of each test case contains one integer $n$ ( $1 \le n \le 10^5$ ) — the length of strings $A$ and $B$ .
The second line of each test case contains string $A$ ( $|A|=n$ ).
The third line of each test case contains string $B$ ( $|B|=n$ ).
Both strings consists of the first $20$ lowercase English alphabet letters (ie. from a to t).
It is guaranteed that the sum of $n$ over all test cases does not exceed $10^5$ .
输出格式
For each test case:
Print on a single line the smallest number of moves she has to do to make strings equal to each other ( $A = B$ ) or $-1$ if there is no way to make them equal.
Print on a single line the smallest number of moves she has to do to make strings equal to each other ( $A = B$ ) or $-1$ if there is no way to make them equal.
输入输出样例
输入 #1
5 3 aab bcc 4 cabc abcb 3 abc tsr 4 aabd cccd 5 abcbd bcdda
输出 #1
2 -1 3 2 -1
说明/提示
- In the $1$ -st test case Koa:
1. selects positions $1$ and $2$ and sets $A_1 = A_2 = $ b ( $\color{red}{aa}b \rightarrow \color{blue}{bb}b$ ).
2. selects positions $2$ and $3$ and sets $A_2 = A_3 = $ c ( $b\color{red}{bb} \rightarrow b\color{blue}{cc}$ ).
- In the $2$ -nd test case Koa has no way to make string $A$ equal $B$ .
- In the $3$ -rd test case Koa:
1. selects position $1$ and sets $A_1 = $ t ( $\color{red}{a}bc \rightarrow \color{blue}{t}bc$ ).
2. selects position $2$ and sets $A_2 = $ s ( $t\color{red}{b}c \rightarrow t\color{blue}{s}c$ ).
3. selects position $3$ and sets $A_3 = $ r ( $ts\color{red}{c} \rightarrow ts\color{blue}{r}$ ).
1. selects positions $1$ and $2$ and sets $A_1 = A_2 = $ b ( $\color{red}{aa}b \rightarrow \color{blue}{bb}b$ ).
2. selects positions $2$ and $3$ and sets $A_2 = A_3 = $ c ( $b\color{red}{bb} \rightarrow b\color{blue}{cc}$ ).
- In the $2$ -nd test case Koa has no way to make string $A$ equal $B$ .
- In the $3$ -rd test case Koa:
1. selects position $1$ and sets $A_1 = $ t ( $\color{red}{a}bc \rightarrow \color{blue}{t}bc$ ).
2. selects position $2$ and sets $A_2 = $ s ( $t\color{red}{b}c \rightarrow t\color{blue}{s}c$ ).
3. selects position $3$ and sets $A_3 = $ r ( $ts\color{red}{c} \rightarrow ts\color{blue}{r}$ ).