【CF207C3】Game with Two Trees
【CF207C3】Game with Two Trees
Description
有两颗动态加入点的树，每条边有一个字符
每次加入完点后T1中到根的链在T2中的匹配次数之和
Input
一行一个数\(q\)
Output
输出\(q\)行每行一个数表示答案
Sample Input
5
1 1 a
2 1 a
1 2 b
2 1 b
2 3 a
Sample Output
1
3
3
4
7
Data Constraint
\(1\le q\le 10^5\)
Solution
给出一个无脑做法
先对T2建广义SAM，再做DAG链剖分
然后T1中每个点可以在DAG链上倍增+Hash找到对应点
接下来写两颗树状数组
分别支持区间加单点查和区间查单点加
复杂度\(O(q\log^2q)\)
Code
#include<bits/stdc++.h>
using namespace std;
#define Fo(i,a,b) for(int i=a;i<=b;i++)
#define Fd(i,a,b) for(int i=a;i>=b;i--)
#define mo 998244353
#define N 100010
#define LL long long
#define ls x<<1
#define rs (x<<1)|1
namespace IO{
const int sz=1<<22;
char a[sz+5],b[sz+5],*p1=a,*p2=a,*t=b,p[105];
inline char gc(){
return p1==p2?(p2=(p1=a)+fread(a,1,sz,stdin),p1==p2?EOF:*p1++):*p1++;
}
template<class T> void read(T& x){
x=0; char c=gc();
for(;c<'0'||c>'9';c=gc());
for(;c>='0'&&c<='9';c=gc())
x=x*10+(c-'0');
}
inline void flush(){fwrite(b,1,t-b,stdout),t=b; }
inline void pc(char x){*t++=x; if(t-b==sz) flush(); }
template<class T> void write(T x,char c='\n'){
if(x<0) pc('-'), x=-x;
if(x==0) pc('0'); int t=0;
for(;x;x/=10) p[++t]=x%10+'0';
for(;t;--t) pc(p[t]); pc(c);
}
struct F{~F(){flush();}}f;
}
using IO::read;
using IO::write;
using IO::gc;
int mod(int x){return x>=mo?x-mo:x;}
int pw[N*2],ipw[N*2];
int n1,n2,id[N*2];
struct node{int ty,u;char ch;}ask[N*2];
vector<int>e1[N*2],e2[N*2];
int num1[N*2],num2[N*2];
struct Trie{
int tot,ch[N*2][26];
int insert(int x,int tr){
ch[x][tr]?x=ch[x][tr]:x=ch[x][tr]=++tot;
return x;
}
void build(int u,int p){
id[u]=p;
for(auto v:e2[u])build(v,insert(p,num2[v]));
}
}T;
struct SAM{
vector<int>e[N*2];
LL s1[N*2],s2[N*2];
int ch[N*2][26],link[N*2],len[N*2],u,rt,tot,lst[N*2];
int nxt[N*2],pre[N*2],son[N*2],rk[N*2],top[N*2],wh[N*2],num;
int line[N*2],cnt,d[N*2];
int g[N*2],pos[N*2];
int *tg[N*2],*tp[N*2];
int vis[N*2],sz[N*2];
int dfn[N*2],siz[N*2],ta;
int extend(int v,int p){
int u=++tot;
for(;v&&!ch[v][p];v=link[v])ch[v][p]=u,len[u]=max(len[u],len[v]+1);
if(!v)link[u]=rt;
else if(len[ch[v][p]]==len[v]+1)link[u]=ch[v][p];
else{
int clone=++tot,Old=ch[v][p];
len[clone]=len[v]+1;
Fo(j,0,25)ch[clone][j]=ch[Old][j];
link[clone]=link[Old];
link[Old]=link[u]=clone;
for(;v&&ch[v][p]==Old;v=link[v])ch[v][p]=clone;
}
return u;
}
void build(){
rt=lst[1]=tot=1;
queue<int>q;
q.push(1);
while(!q.empty()){
int u=q.front();q.pop();
Fo(i,0,25)if(T.ch[u][i]){
int v=T.ch[u][i];
lst[v]=extend(lst[u],i);
q.push(v);
}
}
Fo(i,1,tot)e[link[i]].push_back(i);
}
void dfs(int u){
dfn[u]=++ta;siz[u]=1;
for(auto v:e[u])dfs(v),siz[u]+=siz[v];
}
void DAG(){
queue<int>q;
Fo(i,1,tot) Fo(j,0,25)if(ch[i][j])d[ch[i][j]]++;
q.push(1);
while(!q.empty()){
int u=q.front();q.pop();
line[++cnt]=u;
Fo(i,0,25)if(ch[u][i]){
d[ch[u][i]]--;
if(!d[ch[u][i]])q.push(ch[u][i]);
}
}
Fd(i,cnt,1){
int u=line[i];s1[u]++;
Fo(j,0,25)if(ch[u][j]){
if(s1[ch[u][j]]>s1[nxt[u]])nxt[u]=ch[u][j];
s1[u]+=s1[ch[u][j]];
}
}
Fo(i,1,cnt){
int u=line[i];s2[u]++;
Fo(j,0,25)if(ch[u][j]){
if(s2[u]>s2[pre[ch[u][j]]])pre[ch[u][j]]=u;
s2[ch[u][j]]+=s2[u];
}
}
Fo(i,1,tot) Fo(j,0,25)if(ch[i][j]){
if(nxt[i]==ch[i][j]&&pre[ch[i][j]]==i)son[i]=ch[i][j],wh[i]=j+1;
}
int tmp=0;
Fo(i,1,tot){
int u=line[i];
if(vis[u])continue;
int v=u;
sz[u]=1;rk[u]=++num;top[u]=u;
while(son[v])sz[u]++,v=son[v],rk[v]=++num,top[v]=u;
tg[u]=g+tmp;
tp[u]=pos+tmp;
v=u;
int val=0;
Fo(j,1,sz[u]){
vis[v]=1;
tg[u][j-1]=val;
tp[u][j-1]=v;
val=mod(val+1ll*pw[j-1]*wh[v]%mo);
v=son[v];
}
tmp+=sz[u];
}
}
}S;
struct BIT{
int sum[N*2];
int lowbit(int x){return -x&x;}
void change(int x,int y){for(;x<=S.tot;x+=lowbit(x))sum[x]+=y;}
int query(int x){int res=0;for(;x;x-=lowbit(x))res+=sum[x];return res;}
}T1,T2;
int dep[N*2],fa[N*2][20],val[N*2],q;
LL ans;
int main(){
pw[0]=ipw[0]=1;
Fo(i,1,N*2-10){
pw[i]=37ll*pw[i-1]%mo;
ipw[i]=242816194ll*ipw[i-1]%mo;
}
n1=n2=1;
read(q);
Fo(i,1,q){
int ty,u;char s[3];
read(ty);read(u);s[1]=gc();
ask[i]=(node){ty,ty==1?n1+1:n2+1,s[1]};
if(ty==1){
n1++;
fa[n1][0]=u;
dep[n1]=dep[u]+1;
e1[u].push_back(n1);
num1[n1]=s[1]-'a';
}
if(ty==2){
n2++;
e2[u].push_back(n2);
num2[n2]=s[1]-'a';
}
}
T.tot=1;
T.build(1,1);
S.build();
S.DAG();
S.dfs(1);
Fo(i,1,n1) Fo(j,0,18)fa[i][j+1]=fa[fa[i][j]][j];
Fo(i,2,n1)val[i]=mod(37ll*val[fa[i][0]]%mo+num1[i]+1);
T1.change(1,1);T1.change(S.tot+1,-1);
T2.change(1,1);
ans=1;
Fo(i,1,q){
if(ask[i].ty==1){
int u=1,wh=ask[i].u,flag=1;
while(wh!=1){
int up=S.top[u],lu=S.rk[u]-S.rk[up];
int res=min(dep[wh],S.sz[up]-lu),tmp=lu;
Fd(j,19,0)if(res>=(1<<j)){
if(1ll*mod(val[wh]-1ll*val[fa[wh][j]]*pw[1<<j]%mo+mo)*pw[tmp]%mo==mod(S.tg[up][tmp+(1<<j)]-S.tg[up][tmp]+mo)){
wh=fa[wh][j];tmp+=1<<j;res-=1<<j;
}
}
u=S.tp[up][tmp];
if(wh!=1){
if(!S.ch[u][num1[wh]]){flag=0;break;}
u=S.ch[u][num1[wh]];
wh=fa[wh][0];
}
}
if(flag){
ans+=T2.query(S.dfn[u]+S.siz[u]-1)-T2.query(S.dfn[u]-1);
T1.change(S.dfn[u],1);T1.change(S.dfn[u]+S.siz[u],-1);
}
}else{
ans+=T1.query(S.dfn[S.lst[id[ask[i].u]]]);
T2.change(S.dfn[S.lst[id[ask[i].u]]],1);
}
write(ans);
}
return 0;
}
