[BZOJ 3676] [Apio2014] echo string, bzoj3676

[BZOJ 3676] [Apio2014] echo string, bzoj3676
3676: [Apio2014] Reply string

Time Limit: 20 Sec Memory Limit: 128 MB
Submit: 646 Solved: 219
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Description

Consider a string s that only contains lowercase Latin letters. We define the output of a substring t of s.
The present value is the number of occurrences of t in s multiplied by the length of t. Please find the most
Large value.

Input

Enter only one line, which is a non-empty string s that only contains lowercase letters (a-z.

Output

Output an integer that indicates the maximum occurrence value of the input string.

Sample Input

[Example input l]

Abacaba

[Example input 2]

Www

Sample Output

[Sample output l]

7

[Sample output 2]

4

HINT

A string is a back-to-text string. If it is read from left to right, it is identical to read from right to left.

In the first example, there are seven input strings: a, B, c, aba, aca, bacab, and abacaba:

● A appears four times, and its appearance value is = 4

● B appears twice, and its appearance value is = 2

● C appears once, and its appearance value is l: 1: l = l

● Aba appears twice, and its appearance value is = 6

● Aca appears once, and its appearance value is 1 = 3

● Once bacab appears, its appearance value is = 5

● Abacaba appears once, and its appearance value is = 7

Therefore, the maximum return substring value is 7.

[Data scale and score]

The data must be 1 ≤ string length ≤ 300000.

Manacher + suffix array ~

You can know the process of calculating the length of the input string based on manacher. The length is N Only N Different types of input strings.
(Because only when maxid grows, it will generate different strings)

We only need O (n) The number of occurrences of each string.

Use the suffix array + binary to calculate the number of occurrences.

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <cstdlib>#define M 600010#define LL long longusing namespace std;int n,wa[M],sa[M],he[M],rk[M],f[M/2][20],c[M],wb[M],a[M],b[M];int wv[M],p[M];char s[M];int cmp(int *r,int a,int b,int l){    return r[a]==r[b]&&r[a+l]==r[b+l];}void da(int *r,int *sa,int n,int m){    int *x=wa,*y=wb,*t,i,j,p;    for (i=0;i<m;i++)        c[i]=0;    for (i=0;i<n;i++)        c[x[i]=r[i]]++;    for (i=1;i<=m;i++)        c[i]+=c[i-1];    for (i=n-1;i>=0;i--)        sa[--c[x[i]]]=i;    for (j=1,p=1;p<n;j*=2,m=p)    {        for (p=0,i=n-j;i<n;i++)            y[p++]=i;        for (i=0;i<n;i++)            if (sa[i]>=j)                y[p++]=sa[i]-j;        for (i=0;i<n;i++)            wv[i]=x[y[i]];        for (i=0;i<m;i++)            c[i]=0;        for (i=0;i<n;i++)            c[wv[i]]++;        for (i=1;i<m;i++)            c[i]+=c[i-1];        for (i=n-1;i>=0;i--)            sa[--c[wv[i]]]=y[i];        for (t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;    }}void Calheight(int *r,int *sa,int n){    for (int i=1;i<=n;i++)        rk[sa[i]]=i;    int k=0;    for (int i=0;i<n;i++)    {        if (k) k--;        int j=sa[rk[i]-1];        while (r[i+k]==r[j+k])            k++;        he[rk[i]]=k;    }}void Getst(){    for (int i=0;i<=n;i++)        f[i][0]=he[i];    for (int j=1;(1<<j)<=n;j++)        for (int i=0;i+(1<<j)-1<=n;i++)            f[i][j]=min(f[i][j-1],f[i+(1<<(j-1))][j-1]);}int Calmin(int l,int r){    int len=r-l+1,k=0,now=1;    for (int i=0;i<=20;i++)    {        if (now*2>=len)        {            k=i;            break;        }        now<<=1;    }    return min(f[l][k],f[r-(1<<k)+1][k]);}int Query(int x,int k){    int l,r,ql,qr;    if (he[x+1]<k) qr=x;    else    {        l=x+1,r=n;        while (l<=r)        {            int mid=(l+r)>>1;            if (Calmin(x+1,mid)>=k)                qr=mid,l=mid+1;            else r=mid-1;        }    }    if (he[x]<k) ql=x;    else    {        l=1,r=x-1;        while (l<=r)        {            int mid=(l+r)>>1;            if (Calmin(mid+1,x)>=k)                ql=mid,r=mid-1;            else l=mid+1;        }    }    return qr-ql+1;}int main(){    scanf("%s",s);    n=strlen(s);    for (int i=0;i<n;i++)        a[i]=s[i]-'a'+1;    a[n]=0;    da(a,sa,n+1,27);    Calheight(a,sa,n);    Getst();    for (int i=0;i<n;i++)        b[i<<1]=a[i];    LL ans=1;    int id=0,maxid=-1;    for (int i=0;i<(n<<1);i++)    {        if (maxid>i)            p[i]=min(maxid-i,p[2*id-i]);        else p[i]=1;        while (i-p[i]>=0&&b[i+p[i]]==b[i-p[i]])            p[i]++;        p[i]--;        if (p[i]+i>maxid)        {            for (int j=maxid+1;j<=p[i]+i;j++)                if ((j&1)==0)                 {                    int f=i*2-j;                    int len=(j-f+2)/2;                    ans=max(ans,1LL*len*Query(rk[f>>1],len));                }            maxid=p[i]+i,id=i;        }    }    cout<<ans<<endl;    return 0;}