[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=200010,P1=29,P2=998244353,P3=31,P4=1e9+7;
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-1]*pw1[r-l+1]%P2+P2)%P2; }
int cal2(int l,int r,int hs[]){ return (hs[r]-1ll*hs[l-1]*pw2[r-l+1]%P4+P4)%P4; }

void manacher(char a[],int p[],int hs1[],int hs2[]){
    s[0]='$'; s[1]='#'; int m=2*n+1,id=0,mxl=0;
    rep(i,1,n) s[2*i]=a[i],s[2*i+1]='#';
    rep(i,1,m){
        p[i]=(i<mxl) ? min(p[id*2-i],mxl-i) : 1;
        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]-1);
    }
    rep(i,1,n){
        hs1[i]=(1ll*hs1[i-1]*P1+a[i]-'A')%P2;
        hs2[i]=(1ll*hs2[i-1]*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,1,n){
        int t=pa[2*i]/2,L=t+1,R=min(i,n-i+1);
        while (L<=R){
            int mid=(L+R)>>1;
            if (cal1(i-mid+1,i-t,hsa1)==cal1(n-(i+mid-2)+1,n-(i+t-1)+1,hsb1) && cal2(i-mid+1,i-t,hsa2)==cal2(n-(i+mid-2)+1,n-(i+t-1)+1,hsb2))
                ans=max(ans,mid*2-1),L=mid+1; else R=mid-1;
        }
        if (i==n) continue;
        t=pa[2*i+1]/2,L=t+1,R=min(i,n-i);
        while (L<=R){
            int mid=(L+R)>>1;
            if (cal1(i-mid+1,i-t,hsa1)==cal1(n-(i+mid-1)+1,n-(i+t)+1,hsb1) && cal2(i-mid+1,i-t,hsa2)==cal2(n-(i+mid-1)+1,n-(i+t)+1,hsb2))
                ans=max(ans,mid*2),L=mid+1; else R=mid-1;
        }
    }
}

int main(){
    freopen("palindrome.in","r",stdin);
    freopen("palindrome.out","w",stdout);
    scanf("%d%s%s",&n,A+1,B+1); reverse(B+1,B+n+1);
    pw1[0]=pw2[0]=1;
    rep(i,1,n) pw1[i]=1ll*pw1[i-1]*P1%P2,pw2[i]=1ll*pw2[i-1]*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
",ans);
    return 0;
}