String Transformation 1

String Transformation 1

题目链接：Codeforces Round #659(div2) C.String Transformation 1

题目描述

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 p1,p2,…,pk (k≥1;1≤pi≤n;pi≠pj if i≠j) of A such that Ap1=Ap2=…=Apk=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 p1,p2,…,pk to letter y. More formally: for each i (1≤i≤k) Koa sets Api=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!

Input

Each test contains multiple test cases. The first line contains t (1≤t≤10) — the number of test cases. Description of the test cases follows.

The first line of each test case contains one integer n (1≤n≤105) — 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 10₅.

Output

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.

Example

input

5
3
aab
bcc
4
cabc
abcb
3
abc
tsr
4
aabd
cccd
5
abcbd
bcdda

output

2
-1
3
2
-1

题目大意

修改字符串A若干次，使A=B。在一次操作中，A中相同的字符x可以同时修改为字符y，y>x。求最少操作次数。

解题思路

对于A中的每个字符x，在B中都有对应的字符y，我们的目标是把x变成y。若x=y，则不用修改。对于需要修改的字符，因为字符只能增大，所以我们从a开始修改。对所有需要修改的某个字符x，我们都将它改为最小的y，设为y0，这样没有匹配的将在修改y0时再次修改，直到匹配，不需要多余的操作。

Code

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;

#define MAXN 100005

char a[MAXN], b[MAXN];
vector<char> v[22];

int main(void)
{
    int T; scanf("%d",&T);
    while(T--){
        for(int i=0;i<20;++i) v[i].clear();
        int n; scanf("%d",&n);
        scanf("%s%s",a,b);
        bool flag = true;
        for(int i=0;i<n;++i){
            if(a[i]>b[i]){
                printf("-1
");
                flag = false;
                break;
            }
            if(a[i]==b[i]) continue;
            int k = a[i]-'a';
            v[k].push_back(b[i]);
        }
        if(!flag) continue;
        int cnt = 0;
        for(int i=0;i<20;++i){
            sort(v[i].begin(),v[i].end());
            int num = v[i].size();
            if(num==0) continue;
            char y = v[i][0]; int k = y-'a';
            for(int j=0;j<num;++j){
                if(v[i][j]!=y) v[k].push_back(v[i][j]);
            }
            ++cnt;
        }
        printf("%d
",cnt);
    }
    return 0;
}