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 ;
 }