Educational Codeforces Round 131 (Rated for Div. 2)E

编辑距离与文本转换

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该博客探讨了一种计算从字符串s转换到字符串t所需的最小编辑距离的方法。在这个问题中，允许的操作包括向左移动光标、向右移动、移到开头、移到结尾以及删除字符。给出每种操作的代价，任务是找到从s转换到t的最短路径。通过动态规划，可以找到最长的公共子序列，并计算所需的编辑步骤。对于无法通过这些操作实现的转换，答案为-1。该算法对于文本编辑和字符串比较等场景具有实际应用价值。

E. Text Editor

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You wanted to write a text tt consisting of mm lowercase Latin letters. But instead, you have written a text ss consisting of nn lowercase Latin letters, and now you want to fix it by obtaining the text tt from the text ss.

Initially, the cursor of your text editor is at the end of the text ss (after its last character). In one move, you can do one of the following actions:

press the "left" button, so the cursor is moved to the left by one position (or does nothing if it is pointing at the beginning of the text, i. e. before its first character);
press the "right" button, so the cursor is moved to the right by one position (or does nothing if it is pointing at the end of the text, i. e. after its last character);
press the "home" button, so the cursor is moved to the beginning of the text (before the first character of the text);
press the "end" button, so the cursor is moved to the end of the text (after the last character of the text);
press the "backspace" button, so the character before the cursor is removed from the text (if there is no such character, nothing happens).

Your task is to calculate the minimum number of moves required to obtain the text tt from the text ss using the given set of actions, or determine it is impossible to obtain the text tt from the text ss.

You have to answer TT independent test cases.

Input

The first line of the input contains one integer TT (1≤T≤50001≤T≤5000) — the number of test cases. Then TT test cases follow.

The first line of the test case contains two integers nn and mm (1≤m≤n≤50001≤m≤n≤5000) — the length of ss and the length of tt, respectively.

The second line of the test case contains the string ss consisting of nn lowercase Latin letters.

The third line of the test case contains the string tt consisting of mm lowercase Latin letters.

It is guaranteed that the sum of nn over all test cases does not exceed 50005000 (∑n≤5000∑n≤5000).

Output

For each test case, print one integer — the minimum number of moves required to obtain the text tt from the text ss using the given set of actions, or -1 if it is impossible to obtain the text tt from the text ss in the given test case.

Example

input

Copy

6
9 4
aaaaaaaaa
aaaa
7 3
abacaba
aaa
5 4
aabcd
abcd
4 2
abba
bb
6 4
baraka
baka
8 7
question
problem

output

Copy

多次left和多次right以及多次back和一次home,枚举home的位置即可，最后看看中间最长的连续子串即可。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#define int long long
#define lowbit(x) x&(-x)
#define mp make_pair
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define per(i,n,x) for(int i=n;i>=x;i--)
#define pii pair<int,int>
using namespace std;
const int INF=1e9;
const int mod=998244353;
const int maxn=5e3+5 ;
inline int read()
{
    int x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9')
    {
        if(c=='-') f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9')
    {
        x=x*10+(c-'0');
        c=getchar();
    }
    return x*f;
}
int T;
int a[maxn];
char c[maxn],s[maxn];
int f[maxn][maxn],pre[maxn],suf[maxn];
signed main()
{
    T=read();
    while(T--)
    {
        int ans=INF;
        int n=read(),m=read();
        scanf("%s",c+1);
        scanf("%s",s+1);
        rep(i,1,n) rep(j,1,m) f[i][j]=0;
        int pos=1;
        rep(i,1,n)
        {
            if(s[pos]==c[i]) pos++;
            pre[i]=pos-1;
        }
        if(pre[n]!=m)
        {
            cout<<"-1"<<endl;
            continue;
        }
        pos=m;
        per(i,n,1)
        {
            if(s[pos]==c[i]) pos--;
            suf[i]=m-pos;
        }
        pre[0]=suf[n+1]=0;
        rep(i,1,n)
        {
            rep(j,1,m)
            {
                if(c[i]==s[j]) f[i][j]=f[i-1][j-1]+1;
                if(pre[i-f[i][j]]<j-f[i][j]||suf[i+1]<m-j) continue;
                ans=min(ans,(n-i)+(2*(i-f[i][j])-(j-f[i][j]))+1);
                if(i==f[i][j]) ans=min(ans,n-f[i][j]);
            }
        }
        cout<<min(ans,n)<<endl;
    }
}