CF1383A String Transformation 1 （并查集）

codeforces链接:https://codeforces.com/problemset/problem/1383/A

CF1383A String Transformation 1

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!

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 $ .

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.

输入输出样例 #1

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} $ ).

思路：

首先这道题目是A转换成B，同时改变时候只能由字母表前面位置的字母转换成位置靠后的字母，不难发现，无解的情况是，A中相同位置的字母出现大于B中相同位置的字母，那么正常情况下，我们可以使用并查集进行维护，只有Ai的父节点不等于Bi的父节点时候，进行建边，同时操作数++

题解

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef long long ll;
string s1,s2;
int t;
int ans,len;
vector<int> father = vector<int> (26, 0);


void init() {
    for (int i = 0; i < 26; ++i) {
        father[i] = i;
    }
}

int find(int u) {
    return u == father[u] ? u : father[u]=find(father[u]);
}


void join(int u, int v) {
    u = find(u);
    v = find(v);
    if (u == v) return ;
    ans++;
    father[v] = u;
}


int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin>>t;
    while(t--)
    {
        int cmd=0;
        ans=0;
        cin>>len;
        cin>>s1>>s2;
        init();
        for(int i=0;i<len;i++)
        {

            if(s1[i]>s2[i])
            {
                cmd=1;
                break;
            }
            join(s1[i]-'a',s2[i]-'a');
        }
        if(cmd)
        {
            cout<<-1<<endl;
        }
        else
        {
            cout<<ans<<endl;
        }
    }

    return 0;
}