[BZOJ4755][JSOI2016]扭动的回文串(manacher+Hash)

前两种情况显然直接manacher，对于第三种，枚举回文中心，二分回文半径，哈希判断即可。

 #include<cstdio>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=,P1=,P2=,P3=,P4=1e9+;
 char A[N],B[N],s[N];
 int n,ans,pw1[N],pw2[N],hsa1[N],hsa2[N],hsb1[N],hsb2[N],pa[N],pb[N];

 int cal1(int l,int r,int hs[]){ return (hs[r]-1ll*hs[l-]*pw1[r-l+]%P2+P2)%P2; }
 int cal2(int l,int r,int hs[]){ return (hs[r]-1ll*hs[l-]*pw2[r-l+]%P4+P4)%P4; }

 void manacher(char a[],int p[],int hs1[],int hs2[]){
     s[]='$'; s[]='#'; int m=*n+,id=,mxl=;
     rep(i,,n) s[*i]=a[i],s[*i+]='#';
     rep(i,,m){
         p[i]=(i<mxl) ? min(p[id*-i],mxl-i) : ;
         while (s[i+p[i]]==s[i-p[i]]) p[i]++;
         if (p[i]+i>mxl) mxl=p[i]+i,id=i;
         ans=max(ans,p[i]-);
     }
     rep(i,,n){
         hs1[i]=(1ll*hs1[i-]*P1+a[i]-'A')%P2;
         hs2[i]=(1ll*hs2[i-]*P3+a[i]-'A')%P4;
     }
 }

 void work(char a[],int pa[],int hsa1[],int hsa2[],char b[],int pb[],int hsb1[],int hsb2[]){
     rep(i,,n){
         int t=pa[*i]/,L=t+,R=min(i,n-i+);
         while (L<=R){
             int mid=(L+R)>>;
             if (cal1(i-mid+,i-t,hsa1)==cal1(n-(i+mid-)+,n-(i+t-)+,hsb1) && cal2(i-mid+,i-t,hsa2)==cal2(n-(i+mid-)+,n-(i+t-)+,hsb2))
                 ans=max(ans,mid*-),L=mid+; else R=mid-;
         }
         if (i==n) continue;
         t=pa[*i+]/,L=t+,R=min(i,n-i);
         while (L<=R){
             int mid=(L+R)>>;
             if (cal1(i-mid+,i-t,hsa1)==cal1(n-(i+mid-)+,n-(i+t)+,hsb1) && cal2(i-mid+,i-t,hsa2)==cal2(n-(i+mid-)+,n-(i+t)+,hsb2))
                 ans=max(ans,mid*),L=mid+; else R=mid-;
         }
     }
 }

 int main(){
     freopen("palindrome.in","r",stdin);
     freopen("palindrome.out","w",stdout);
     scanf("%d%s%s",&n,A+,B+); reverse(B+,B+n+);
     pw1[]=pw2[]=;
     rep(i,,n) pw1[i]=1ll*pw1[i-]*P1%P2,pw2[i]=1ll*pw2[i-]*P3%P4;
     manacher(A,pa,hsa1,hsa2); manacher(B,pb,hsb1,hsb2);
     work(A,pa,hsa1,hsa2,B,pb,hsb1,hsb2);
     work(B,pb,hsb1,hsb2,A,pa,hsa1,hsa2);
     printf("%d\n",ans);
     return ;
 }