http://codeforces.com/gym/100623/attachments E題
第一個優化
它雖然是鏡像對稱，但它畢竟是一一對稱的，所以可以匹配串和模式串都從頭到尾顛倒一下
第二個優化，與次數無關，所以排個序就完事了

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<ctime>
 7 #include<set>
 8
 
 #include<map>
 9 #include<stack>
10 #include<cstring>
11 #define inf 2147483647
12 #define ls rt<<1
13 #define rs rt<<1|1
14 #define lson ls,nl,mid,l,r
15 #define rson rs,mid+1,nr,l,r
16 #define N 100010
17 #define For(i,a,b) for(int i=a;i<=b;i++)
18 #define p(a) putchar(a)
19
 
 #define g() getchar()
20 
21 using namespace std;
22 int n;
23 char a[110],b[110],c[110],d[110];
24 
25 struct node{
26     int f;
27     int s;
28     bool operator < (const node & temp)const{
29         if(f==temp.f)
30             return s<temp.s;
31         return f<temp.f;
 
32     }
33 }a1[110],a2[110];
34 
35 void in(int &x){
36     int y=1;
37     char c=g();x=0;
38     while(c<‘0‘||c>‘9‘){
39         if(c==‘-‘)y=-1;
40         c=g();
41     }
42     while(c<=‘9‘&&c>=‘0‘){
43         x=(x<<1)+(x<<3)+c-‘0‘;c=g();
44     }
45     x*=y;
46 }
47 void o(int x){
48     if(x<0){
49         p(‘-‘);
50         x=-x;
51     }
52     if(x>9)o(x/10);
53     p(x%10+‘0‘);
54 }
55 int main(){
56     freopen("enchanted.in","r",stdin);
57     freopen("enchanted.out","w",stdout);
58     cin>>(a+1)>>(b+1)>>(c+1)>>(d+1);
59     n=strlen(a+1);
60     For(i,1,n){
61         a1[i].f=a[i]-‘A‘+1;
62         a1[i].s=b[n-i+1]-‘A‘+1;
63     }
64     For(i,1,n){
65         a2[i].f=c[i]-‘A‘+1;
66         a2[i].s=d[n-i+1]-‘A‘+1;
67     }
68     sort(a1+1,a1+n+1);
69     sort(a2+1,a2+n+1);
70     For(i,1,n)
71         if(a1[i].f!=a2[i].f||a1[i].s!=a2[i].s){
72             cout<<"No";
73             return 0;
74         }
75     cout<<"Yes";
76     return 0;
77 }

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