http://codeforces.com/gym/100623/attachments E题

http://codeforces.com/gym/100623/attachments E题
第一个优化
它虽然是镜像对称，但它毕竟是一一对称的，所以可以匹配串和模式串都从头到尾颠倒一下
第二个优化，与次数无关，所以排个序就完事了

 #include<iostream>

 #include<cstdio>

 #include<queue>

 #include<algorithm>

 #include<cmath>

 #include<ctime>

 #include<set>

 #include<map>

 #include<stack>

 #include<cstring>

 #define inf 2147483647

 #define ls rt<<1

 #define rs rt<<1|1

 #define lson ls,nl,mid,l,r

 #define rson rs,mid+1,nr,l,r

 #define N 100010

 #define For(i,a,b) for(int i=a;i<=b;i++)

 #define p(a) putchar(a)

 #define g() getchar()

 using namespace std;

 int n;

 char a[],b[],c[],d[];

 struct node{

     int f;

     int s;

     bool operator < (const node & temp)const{

         if(f==temp.f)

             return s<temp.s;

         return f<temp.f;

     }

 }a1[],a2[];

 void in(int &x){

     int y=;

     char c=g();x=;

     while(c<''||c>''){

         if(c=='-')y=-;

         c=g();

     }

     while(c<=''&&c>=''){

         x=(x<<)+(x<<)+c-'';c=g();

     }

     x*=y;

 }

 void o(int x){

     if(x<){

         p('-');

         x=-x;

     }

     if(x>)o(x/);

     p(x%+'');

 }

 int main(){

     freopen("enchanted.in","r",stdin);

     freopen("enchanted.out","w",stdout);

     cin>>(a+)>>(b+)>>(c+)>>(d+);

     n=strlen(a+);

     For(i,,n){

         a1[i].f=a[i]-'A'+;

         a1[i].s=b[n-i+]-'A'+;

     }

     For(i,,n){

         a2[i].f=c[i]-'A'+;

         a2[i].s=d[n-i+]-'A'+;

     }

     sort(a1+,a1+n+);

     sort(a2+,a2+n+);

     For(i,,n)

         if(a1[i].f!=a2[i].f||a1[i].s!=a2[i].s){

             cout<<"No";

             return ;

         }

     cout<<"Yes";

     return ;

 }