cf700E. Cool Slogans

题解: 首先建出后缀自动机 我们考虑对于每个节点i 在其parent树上找到离其最近的j让s[j]子串在s[i]中出现>=2次以上(因为在j祖先节点必然都满足条件 但显然最近的最优) 所以把原parent树转化成新树 新树的最大高度即为答案 具体实现 可以倍增+线段树合并

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <set>
#include <map>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define link(x) for(edge *j=h[x];j;j=j->next)
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,r,l) for(int i=r;i>=l;i--)
const int MAXN=4e5+10;
const double eps=1e-8;
#define ll long long
using namespace std;
struct edge{int t;edge*next;}e[MAXN<<1],*h[MAXN],*o=e;
void add(int x,int y){o->t=y;o->next=h[x];h[x]=o++;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}



int sz[MAXN],dis[MAXN],fa[MAXN],ch[MAXN][26];
int rt,cur,cnt,Rt[MAXN],n,cnt1;
vector<int>vec[MAXN];
char str[MAXN];
typedef struct node{
    int l,r,num;
}node;
node d[MAXN*42];
void update(int &x,int l,int r,int t){
    if(!x){x=++cnt1;d[x].l=0;d[x].r=0;d[x].num=0;}
    if(l==r){d[x].num=1;return ;}
    int mid=(l+r)>>1;
    if(t<=mid)update(d[x].l,l,mid,t);
    else update(d[x].r,mid+1,r,t);
    d[x].num=d[d[x].l].num+d[d[x].r].num;
}
void merge(int &x,int y,int l,int r){
    if(!y)return ;
    if(!x){x=y;return ;}
    int t=++cnt1;d[t]=d[x];d[t].num+=d[y].num;x=t;
    if(l==r)return ;
    int mid=(l+r)>>1;
    merge(d[x].l,d[y].l,l,mid);
    merge(d[x].r,d[y].r,mid+1,r);
}
int ans;
void querty(int x,int l,int r,int ql,int qr){
    if(!x)return ;
    if(ql<=l&&r<=qr){ans+=d[x].num;return ;}
    int mid=(l+r)>>1;
    if(ql<=mid)querty(d[x].l,l,mid,ql,qr);
    if(qr>mid)querty(d[x].r,mid+1,r,ql,qr);
}

void built(int x){
    int last=cur;cur=++cnt;dis[cur]=dis[last]+1;int p=last;sz[cur]=dis[cur];
    update(Rt[cur],1,n,dis[cur]);
    for(;p&&!ch[p][x];p=fa[p])ch[p][x]=cur;
    if(!p)fa[cur]=rt;
    else{
	int q=ch[p][x];
	if(dis[q]==dis[p]+1)fa[cur]=q;
	else{
	    int nt=++cnt;dis[nt]=dis[p]+1;
	    memcpy(ch[nt],ch[q],sizeof(ch[q]));
	    fa[nt]=fa[q];fa[q]=fa[cur]=nt;
	    for(;ch[p][x]==q;p=fa[p])ch[p][x]=nt;
	}
    }
}
int f[MAXN][21],dep[MAXN];

void dfs(int x,int pre,int deep){
    f[x][0]=pre;dep[x]=deep+1;
  //  cout<<x<<"===="<<pre<<endl;
    link(x){
	dfs(j->t,x,deep+1);
	sz[x]=sz[j->t];
	merge(Rt[x],Rt[j->t],1,n);
    }
 //   cout<<x<<"===="<<Rt[x]<<" "<<d[Rt[x]].num<<endl;
}
int Lca(int x,int t){
    for(int i=0;i<=20;i++)if(t&(1<<i))x=f[x][i];
    return x;
}
bool check(int l,int r,int x,int t){
    int t1=Lca(x,t);
    //cout<<fa[t1]<<" "<<dis[fa[t1]]<<endl;
    ans=0;querty(Rt[t1],1,n,l+dis[fa[t1]],r);
    //cout<<l+dis[fa[t1]]<<" "<<r<<" "<<x<<" "<<t<<" "<<t1<<"::: "<<ans<<endl;
    return (ans>=2);
}

void slove(int x,int t){
    if(!t)return ;
    //cout<<x<<"===="<<t<<endl;
    int l=1;int r=t;int ans1=0;
    //cout<<sz[x]-dis[x]+1<<" "<<sz[x]<<" "<<x<<" "<<t<<endl;
    while(l<=r){
	int mid=(l+r)>>1;
	if(check(sz[x]-dis[x]+1,sz[x],x,mid))ans1=mid,r=mid-1;
	else l=mid+1;
    }
    int t1=Lca(x,ans1);
    if(t1==x)t1=rt;
    vec[t1].pb(x);
  //  cout<<t1<<"-----"<<x<<endl;
}
void dfs1(int x){
    inc(i,1,20)f[x][i]=f[f[x][i-1]][i-1];
    if(x!=rt)slove(x,dep[x]-1);
    link(x){
	dfs1(j->t);
    }
}
int maxx;
void dfs2(int x,int deep){
    maxx=max(maxx,deep);
    for(int i=0;i<vec[x].size();i++){
	dfs2(vec[x][i],deep+1);
    }
}

int main(){
    n=read();scanf("%s",str+1);
    cnt=cur=rt=1;
    inc(i,1,n)built(str[i]-'a');
    inc(i,1,cnt)add(fa[i],i);
    dfs(rt,0,0);dfs1(rt);
  //  cout<<"sb"<<endl;
    maxx=0;dfs2(rt,0);
    printf("%d
",maxx);
}

　　

　　　　　　　　　　　　　　　　　　E. Cool Slogans

　　　　　　　　　　　　　　　　　　time limit per test

　　　　　　　　　　　　　　　　　　4 seconds

　　　　　　　　　　　　　　　　　　memory limit per test

　　　　　　　　　　　　　　　　　　512 megabytes

input

standard input

output

standard output

Bomboslav set up a branding agency and now helps companies to create new logos and advertising slogans. In term of this problems, slogan of the company should be a non-empty substring of its name. For example, if the company name is "hornsandhoofs", then substrings "sand" and "hor" could be its slogans, while strings "e" and "hornss" can not.

Sometimes the company performs rebranding and changes its slogan. Slogan A is considered to be cooler than slogan B if B appears in Aas a substring at least twice (this occurrences are allowed to overlap). For example, slogan A =  "abacaba" is cooler than slogan B = "ba", slogan A =  "abcbcbe" is cooler than slogan B =  "bcb", but slogan A =  "aaaaaa" is not cooler than slogan B =  "aba".

You are given the company name w and your task is to help Bomboslav determine the length of the longest sequence of slogans s1, s2, ..., sk, such that any slogan in the sequence is cooler than the previous one.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the length of the company name that asks Bomboslav to help. The second line contains the string w of length n, that consists of lowercase English letters.

Output

Print a single integer — the maximum possible length of the sequence of slogans of the company named w, such that any slogan in the sequence (except the first one) is cooler than the previous