Description
给定一个初始字符串，要求支持两种操作：在当前字符串后面加上一个字符串；询问一个字符串在当前字符串中作为子串出现了几次。
Solution
LCT 维护后缀自动机，询问的就是后缀树的子树大小
于是我们暴力维护子树大小
调了大半天，最后终于用一个非常 tricky 的办法把它搞定了
首先这是因为我之前写的维护子树和都是假的（实际上只对整个连通块保持正确性，毕竟是从 BJOI2014 大融合 那题引申出来的
将计就计，我们干脆在每次询问的时候把对应子树单独切出来求和，然后再连接回去
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 3000005;
const int N = 3000005;
string decodeWithMask(string s, int mask) {
for (int j = 0; j < s.length(); j++) {
mask = (mask * 131 + j) % s.length();
char t = s[j];
s[j] = s[mask];
s[mask] = t;
}
return s;
}
namespace lct{
int top, q[N], ch[N][2], fa[N], rev[N], val[N];
int sz[N], si[N];
inline void pushup(int x){
sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+si[x]+val[x]; //
}
inline void pushdown(int x){
if(!rev[x]) return;
rev[ch[x][0]]^=1;
rev[ch[x][1]]^=1;
rev[x]^=1;
swap(ch[x][0],ch[x][1]);
}
inline bool isroot(int x){
return ch[fa[x]][0]!=x && ch[fa[x]][1]!=x;
}
inline void rotate(int p){
int q=fa[p], y=fa[q], x=ch[fa[p]][1]==p;
ch[q][x]=ch[p][x^1]; fa[ch[q][x]]=q;
ch[p][x^1]=q; fa[q]=p; fa[p]=y;
if(y) if(ch[y][0]==q) ch[y][0]=p;
else if(ch[y][1]==q) ch[y][1]=p;
pushup(q); pushup(p);
}
inline void splay(int x){
q[top=1]=x;
for(int i=x;!isroot(i);i=fa[i]) q[++top]=fa[i];
for(int i=top;i;i--) pushdown(q[i]);
for(;!isroot(x);rotate(x))
if(!isroot(fa[x]))
rotate((ch[fa[x]][0]==x)==(ch[fa[fa[x]]][0]==fa[x])?fa[x]:x);
}
void access(int x){
for(int t=0;x;t=x,x=fa[x])
splay(x),si[x]+=sz[ch[x][1]],si[x]-=sz[t],ch[x][1]=t,pushup(x); //
}
void makeroot(int x){
access(x);
splay(x);
rev[x]^=1;
}
int find(int x){
access(x);
splay(x);
while(ch[x][0]) x=ch[x][0];
return x;
}
void split(int x,int y){
makeroot(x);
access(y);
splay(y);
}
void cut(int x,int y){
if(!x || !y) return;
split(x,y);
if(ch[y][0]==x)
ch[y][0]=0, fa[x]=0;
pushup(y); //
makeroot(1);
}
void link(int x,int y){
access(y);
splay(y);
fa[x]=y;
si[y]+=sz[x]; //
pushup(y); //
makeroot(1);
}
void print()
{
for(int i=1;i<=10;i++)
{
cout<<" vertex "<<i<<" sz="<<sz[i]<<" si="<<si[i]<<" fa="<<fa[i]<<" ch="<<ch[i][0]<<","<<ch[i][1]<<endl;
}
}
}
namespace sam {
int maxlen[Maxn], trans[Maxn][26], link[Maxn], Size=1, Last=1;
int t[Maxn], a[Maxn], cnt[Maxn], f[Maxn];
inline void extend(int id) {
int cur = (++ Size), p;
lct::val[Size]=1;
lct::pushup(Size);
maxlen[cur] = maxlen[Last] + 1;
cnt[cur] = 1;
for (p = Last; p && !trans[p][id]; p = link[p]) trans[p][id] = cur;
if (!p) link[cur] = 1, lct::link(cur,1);
else {
int q = trans[p][id];
if (maxlen[q] == maxlen[p] + 1) link[cur] = q, lct::link(cur,q);
else {
int clone = (++ Size);
//lct::sz[Size]=1;
maxlen[clone] = maxlen[p] + 1;
for(int i=0;i<26;i++) trans[clone][i] = trans[q][i];
link[clone] = link[q];
lct::link(clone,link[q]);
for (; p && trans[p][id] == q; p = link[p]) trans[p][id] = clone;
if(link[q]) lct::cut(q,link[q]);
link[cur] = link[q] = clone;
lct::link(cur,clone);
lct::link(q,clone);
}
}
Last = cur;
}
void extend(string s)
{
for(char t:s)
{
extend(t-'A');
lct::access(1);
lct::splay(1);
//lct::print();
}
}
int query(string s)
{
int p=1,len=s.length();
for(int i=0;i<len;i++)
{
p=trans[p][s[i]-'A'];
if(!p) return 0;
}
//cout<<"query "<<p<<endl;
if(link[p]) lct::cut(p,link[p]);
lct::makeroot(p);
int ans = lct::sz[p];
if(link[p]) lct::link(p,link[p]);
lct::makeroot(1);
return ans;
}
}
signed main()
{
int n,lastans=0;
cin>>n;
string s;
cin>>s;
sam::extend(s);
string op;
while(n--)
{
cin>>op>>s;
if(op[0]=='A')
{
sam::extend(decodeWithMask(s,lastans));
}
else
{
int tans;
cout<<(tans=sam::query(decodeWithMask(s,lastans)))<<endl;
lastans^=tans;
}
//lct::print();
}
}
/*
20
BBBAABABBA
ADD ABBABABABBBB
QUERY BABABAABBA
QUERY AABABBBAAB
QUERY BABABBABAABBBB
ADD BABAAAAAABABB
ADD BABABBBBABBAAA
ADD BBBAABABAA
QUERY BBBBAAA
ADD BBBAAABBBA
QUERY BABBBBA
QUERY BABABBAAB
ADD BBBBBBBBB
ADD BAABAAB
ADD BAAAAAB
QUERY ABBBBBB
ADD ABBBAAA
QUERY ABABBBAABBABA
QUERY ABAAAABBABBB
ADD AAAAABAAAAA
QUERY ABBAABBBBBB
*/