Thinking
I feel this good fairy ah off-line thinking qwq
For each interrogation \ ([l, r] \ ) is actually required \ (P_ {max} \) , so \ (LCS (S [. 1, P], S [. 1, R & lt])> PL \) , also is \ (LCS (S [. 1, P], S [. 1, R & lt]) + L> P \) .
First, press the interrogation offline \ (R & lt \) sort, then sweep from right to left, each
- It has been added to the list before the process of inquiry, to see whether the current position can be used as \ (the p-\) , and then ask to be deleted has been processed.
- The current inquiry into the go.
Playing out the suffix tree built, then the left side of the inequality is the \ ({dep_ LCA (P, R & lt)} L + \) .
The practice of violence is to enumerate \ (LCA \) , then check when dealing with inquiries of this position \ (l \) , plug ask when the \ (l \) to go into.
Split with chain accelerated tree: For each strand to \ (tag \) hit the deepest place. When asked on check processing \ (p \) to the root of the \ (\ max (dep_x + l ) \) and the \ (tag \) which inquiries from, and then delete it. Plug ask when he jumped all the way up, hit \ (Tag \) .
How the query? Sub-biggest \ (tag \) is discussed above or below, if it is directly above the check, otherwise check \ (\ max (l) \ ) plus the current \ (DEP \) .
Since each point can have a maximum of \ (tag \) in the tree line, so even with a \ (set \) record the point which tags.
(Probably speaks not very clear qwq)
Code
#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define sz 402020
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
char __sr[1<<21],__z[20];int __C=-1,__zz=0;
inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
inline void print(register int x)
{
if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
while(__z[++__zz]=x%10+48,x/=10);
while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
}
void file()
{
#ifdef NTFOrz
freopen("a.in","r",stdin);
#endif
}
inline void chktime()
{
#ifndef ONLINE_JUDGE
cout<<(clock()-t)/1000.0<<'\n';
#endif
}
#ifdef mod
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
ll inv(ll x){return ksm(x,mod-2);}
#else
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
#endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int n,Q;
char s[sz];
int pos[sz];
struct hh{int l,r,id;const bool operator < (const hh &x) const {return r>x.r;}}q[sz];
int ans[sz];
int TT;
//struct SegmentTree
//{
pii tr[sz<<2][2];
#define ls k<<1
#define rs k<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
void pushup(int k){tr[k][TT]=max(tr[ls][TT],tr[rs][TT]);}
void modify(int k,int l,int r,int x,pii w)
{
if (l==r) return (void)(tr[k][TT]=w);
int mid=(l+r)>>1;
if (x<=mid) modify(lson,x,w);
else modify(rson,x,w);
pushup(k);
}
pii query(int k,int l,int r,int x,int y)
{
if (x<=l&&r<=y) return tr[k][TT];
int mid=(l+r)>>1;pii ret=MP(-1e9,0);
if (x<=mid) chkmax(ret,query(lson,x,y));
if (y>mid) chkmax(ret,query(rson,x,y));
return ret;
}
//}T1,T2;
// 1：维护一段里面最大的值是多少，以及它的位置。
// 2：维护一段里面加的l的最大值是多少，以及它的位置。
namespace Tree
{
struct hh{int t,nxt;}edge[sz];
int head[sz],ecnt;
void make_edge(int f,int t){edge[++ecnt]=(hh){t,head[f]};head[f]=ecnt;}
int dfn[sz],top[sz],bot[sz],size[sz],son[sz],fa[sz],len[sz],T;
#define v edge[i].t
void dfs1(int x)
{
size[x]=1;
go(x)
{
fa[v]=x;dfs1(v);
size[x]+=size[v];
if (size[v]>size[son[x]]) son[x]=v;
}
}
void dfs2(int x,int tp)
{
dfn[x]=++T;
top[x]=tp,bot[tp]=x;
if (son[x]) dfs2(son[x],tp);
go(x) if (v!=son[x]) dfs2(v,v);
}
#undef v
set<pii>s[sz];
// 维护这个地方打的标记有哪些
pii query(int x)
{
pii ret=MP(0,0);
while (x)
{
TT=0;pii w1=::query(1,1,T,dfn[top[x]],dfn[x]);
TT=1;pii w2=::query(1,1,T,dfn[x],dfn[bot[top[x]]]);
w2.fir+=len[x];
chkmax(ret,w1),chkmax(ret,w2);
x=fa[top[x]];
}
return ret;
}
void modify(int x,int w,int id,bool add)
{
while (x)
{
if (add) s[x].insert(MP(w,id));
else s[x].erase(MP(w,id));
if (s[x].size())
{
auto cur=*s[x].rbegin();
TT=1;::modify(1,1,T,dfn[x],cur);
cur.fir+=len[x];
TT=0;::modify(1,1,T,dfn[x],cur);
}
else TT=0,::modify(1,1,T,dfn[x],MP(0,0)),TT=1,::modify(1,1,T,dfn[x],MP(0,0));
x=fa[top[x]];
}
}
}
namespace SAM
{
struct hh{int link,len,ch[28];}a[sz];
int lst=1,cnt=1,root=1;
int add(int c)
{
int cur=++cnt,p=lst;lst=cur;a[cur].len=a[p].len+1;
while (p&&!a[p].ch[c]) a[p].ch[c]=cur,p=a[p].link;
if (!p) return a[cur].link=1,cur;
int q=a[p].ch[c];
if (a[q].len==a[p].len+1) return a[cur].link=q,cur;
int t=++cnt;a[t]=a[q];a[q].link=a[cur].link=t;a[t].len=a[p].len+1;
while (p&&a[p].ch[c]==q) a[p].ch[c]=t,p=a[p].link;
return cur;
}
void build(){rep(i,2,cnt) Tree::make_edge(a[i].link,i),Tree::len[i]=a[i].len;}
}
int main()
{
file();
read(n,Q);
cin>>(s+1);
rep(i,1,n) pos[i]=SAM::add(s[i]-'a'+1);
SAM::build();
Tree::dfs1(1),Tree::dfs2(1,1);
rep(i,1,Q) read(q[i].l,q[i].r),q[i].id=i;
sort(q+1,q+Q+1);
int p=1;
drep(i,n,1)
{
while (233)
{
pii w=Tree::query(pos[i]);int id=w.sec;
if (w.fir<=i) break;
ans[q[id].id]=i-q[id].l+1;
Tree::modify(pos[q[id].r],q[id].l,id,0);
}
while (p<=Q&&q[p].r==i) Tree::modify(pos[q[p].r],q[p].l,p,1),++p;
}
rep(i,1,Q) printf("%d\n",ans[i]);
return 0;
}