BZOJ4755 JSOI2016扭动的回文串（二分答案+哈希）

　　显然答案应该是由单串以某位置为中心的极长回文串继续在另一个串里拓展得到的。枚举中间位置二分答案，哈希判断即可。注意考虑清楚怎么处理偶回文，比如像manacher一样加分隔符。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
#define ul unsigned long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,ans;
ul ha[N],rha[N],hb[N],rhb[N],p[N];
char a[N],b[N],s1[N],s2[N];
ul hash(ul *a,int l,int r,int op)
{
    return a[r]-a[l-op]*p[abs(r-l)+1];
}
void solve(ul *a,ul *ra,ul *b,ul *rb,int op,int n)
{
    for (int i=1;i<=n;i++)
    {
        int l=0,r=min(n-i,i-1),len=0;
        while (l<=r)
        {
            int mid=l+r>>1;
            if (hash(a,i-mid,i,1)==hash(ra,i+mid,i,-1)) len=mid,l=mid+1;
            else r=mid-1;
        }
        ans=max(ans,len);
        l=1,r=min(i-len+1,n-i-len);
        while (l<=r)
        {
            int mid=l+r>>1;
            if (op?hash(b,i-len-mid+1,i-len,1)==hash(ra,i+len+mid,i+len+1,-1):hash(a,i-len-mid,i-len-1,1)==hash(rb,i+len+mid-1,i+len,-1)) ans=max(ans,len+mid),l=mid+1;
            else r=mid-1;
        }
    }
}
void work(int n)
{
    n=n*2+2;
    for (int i=1;i<=n;i++) ha[i]=ha[i-1]*509+a[i];
    for (int i=n;i>=1;i--) rha[i]=rha[i+1]*509+a[i];
    for (int i=1;i<=n;i++) hb[i]=hb[i-1]*509+b[i];
    for (int i=n;i>=1;i--) rhb[i]=rhb[i+1]*509+b[i];
    p[0]=1;for (int i=1;i<=n;i++) p[i]=p[i-1]*509;
    solve(ha,rha,hb,rhb,0,n),solve(hb,rhb,ha,rha,1,n);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4755.in","r",stdin);
    freopen("bzoj4755.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read();
    scanf("%s",s1+1),scanf("%s",s2+1);
    for (int i=n;i>=1;i--) a[i*2+2]='$',a[i*2+1]=s1[i];a[2]='$';a[1]='#';
    for (int i=n;i>=1;i--) b[i*2+1]='$',b[i*2]=s2[i];b[1]='$';b[n*2+2]='@';
    work(n);
    for (int i=n;i>=1;i--) a[i*2+1]='$',a[i*2]=s1[i];a[1]='$';a[n*2+2]='@';
    for (int i=n;i>=1;i--) b[i*2+2]='$',b[i*2+1]=s2[i];b[2]='$';b[1]='#';
    work(n);
    cout<<ans;
    return 0;
}