SGU 439 A Secret Book

解法：

　　对于第二个串，循环移动能得到的字典序最小的串，可以直接用最小表示法搞定。

　　然后用最小表示的第二个串和第一个串做两次扩展KMP，一次正常求，另外一次将两个串都反转一下，然后扫一遍ex[]数组

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N = (int)2e6+;
 char a[N],b[N],c[N];
 int next[N],exa[N],exb[N];
 void getnext(char *s){
     int len = strlen(s),cur = ;
     next[] = len;
     while(cur < len&&s[cur]==s[cur+])cur++;
     next[] = cur;cur = ;
     for(int k = ;k<len;k++){
         int p = cur + next[cur] - ,L = next[k-cur];
         if(k + L - >= p){
             int j = (p-k+)>?(p-k+):;
             while(k+j<len&&s[k+j]==s[j])j++;
             next[k] = j;
             cur = k;
         }else
             next[k] = L;
     }
 }
 void exkmp(char *s1,char *s2,int *ex){
     getnext(s2);
     int l1 = strlen(s1),l2 = strlen(s2),cur = ;
     while(cur < min(l1,l2)&&s1[cur]==s2[cur])cur++;
     ex[] = cur;cur = ;
     for(int k = ;k < l1;k++){
         int p = cur + ex[cur] - ,L = next[k-cur];
         if(k + L - >= p){
             int j = (p-k+)>?(p-k+):;
             while(k+j<l1&&j<l2&&s1[k+j]==s2[j])j++;
             ex[k] = j;
             cur = k;
         }else
             ex[k] = L;
     }
 }
 int MinRep(char *s){
     int i = ,j = ,k = ,t,len = strlen(s);
     while(i<len&&j<len&&k<len){
         t = s[(i+k)%len] - s[(j+k)%len];
         if(t==)k++;
         else{
             if(t>)
                 i += k + ;
             else
                 j += k + ;
             if(i==j)j++;
             k = ;
         }
     }
     return min(i,j);
 }
 inline int dis(int now,int sz){
     return min(now,sz-now-);
 }
 int main(){
     int n,m;
     while(~scanf("%d%d",&n,&m)){
         scanf("%s%s",a,b);
         int start = MinRep(b);
         for(int i = ,j = start;i < m;i++,j = (j+)%m)c[i] = b[j];c[m] = ;
         for(int i = n;i < n+n;i++)a[i] = a[i-n];a[n+n] = ;
         exkmp(a,c,exa);
         reverse(a,a+*n);reverse(c,c+m);
         exkmp(a,c,exb);
         reverse(exb,exb+n*);
         reverse(c,c+m);
         reverse(a,a+*n);
         int ans = -;
         for(int i = ;i < n;i++)
             if(exa[i] == m || exa[i] + exb[i+m-] == m-){
                 if(ans==-)ans = i;
                 else if(dis(i,n)<dis(ans,n))ans = i;
             }
         puts(c);
         for(int i = ,j = ans;i < n;i++,j = (j+)%n)putchar(a[j]);puts("");
     }
     return ;
 }