B - New Place

https://atcoder.jp/contests/arc154/tasks/arc154_b

思路

https://blog.csdn.net/nike0good/article/details/128751025

由于操作的特殊性，从s的第一个字符删除，插入后面任意地方，

只需要判定s的尾部后缀子串，在t中顺序对应的情况，

这个后缀子串中的字符顺序已经满足条件，

后缀子串之前的字符使用 规定的操作，可以放置在t中的对应位置。

Code

COPY:  https://blog.csdn.net/nike0good/article/details/128751025

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
int main()
{
//    freopen("B.in","r",stdin);
//    freopen(".out","w",stdout);
    int n=read();
    string s,t;
    cin>>s>>t;
    int c[256]={};
    Rep(i,n) c[s[i]]++,c[t[i]]--;
    Fork(i,'a','z') if(c[i]){
        puts("-1");return 0;
    }
    int p=n-1,ans=0;
    RepD(i,n-1) {
        while(p>=0 && s[i]!=t[p]) --p;
        if(p<0) break;
        --p,++ans;
    }cout<<n-ans;
    
    return 0;
}