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[110],b[110],c[110],d[110];

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[110],a2[110];

void in(int &x){
    int y=1;
    char c=g();x=0;
    while(c<'0'||c>'9'){
        if(c=='-')y=-1;
        c=g();
    }
    while(c<='9'&&c>='0'){
        x=(x<<1)+(x<<3)+c-'0';c=g();
    }
    x*=y;
}
void o(int x){
    if(x<0){
        p('-');
        x=-x;
    }
    if(x>9)o(x/10);
    p(x%10+'0');
}
int main(){
    freopen("enchanted.in","r",stdin);
    freopen("enchanted.out","w",stdout);
    cin>>(a+1)>>(b+1)>>(c+1)>>(d+1);
    n=strlen(a+1);
    For(i,1,n){
        a1[i].f=a[i]-'A'+1;
        a1[i].s=b[n-i+1]-'A'+1;
    }
    For(i,1,n){
        a2[i].f=c[i]-'A'+1;
        a2[i].s=d[n-i+1]-'A'+1;
    }
    sort(a1+1,a1+n+1);
    sort(a2+1,a2+n+1);
    For(i,1,n)
        if(a1[i].f!=a2[i].f||a1[i].s!=a2[i].s){
            cout<<"No";
            return 0;
        }
    cout<<"Yes";
    return 0;
}