[bzoj2555]SubString

题目大意

要求兹瓷两个操作

1、在字符串后插入字符串变为新的字符串

2、询问字符串内某字符串的出现次数

LCT维护SAM

对于询问字符串出现次数而且要兹瓷在尾部添加字符，容易想到是SAM，根据pre指针建树，然后用LCT维护这颗pre树即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef int ll;
const ll maxn=600000+10;
ll father[maxn*2],tree[maxn*2][2],sta[maxn*2],pp[maxn*2];
long long size[maxn*2],add[maxn*2];
ll pre[maxn*2],g[maxn*2][26],step[maxn*2];
char s[maxn];
bool czy;
char ch;
ll i,j,k,l,t,n,m,mask,tot,ans,last,top;
void get_s(ll mask) {
scanf("%s",s+1);
n=strlen(s+1);
fo(i,0,n-1) {
mask=(mask*131+i)%n;
swap(s[i+1],s[mask+1]);
}
}
ll pd(ll x){
return tree[father[x]][1]==x;
}
void clear(ll x){
if (add[x]){
size[tree[x][0]]+=add[x];
add[tree[x][0]]+=add[x];
size[tree[x][1]]+=add[x];
add[tree[x][1]]+=add[x];
add[x]=0;
}
}
void remove(ll x,ll y){
top=0;
while (x!=y){
sta[++top]=x;
x=father[x];
}
while (top){
clear(sta[top]);
top--;
}
}
void rotate(ll x){
ll y=father[x],z=pd(x);
father[x]=father[y];
if (father[y]) tree[father[y]][pd(y)]=x;
tree[y][z]=tree[x][1-z];
if (tree[x][1-z]) father[tree[x][1-z]]=y;
tree[x][1-z]=y;
father[y]=x;
if (pp[y]) pp[x]=pp[y],pp[y]=0;
}
void splay(ll x,ll y){
remove(x,y);
while (father[x]!=y){
if (father[father[x]]!=y)
if (pd(x)==pd(father[x])) rotate(father[x]);else rotate(x);
rotate(x);
}
}
void access(ll x){
ll y;
splay(x,0);
//clear(x);
if (tree[x][1]) pp[tree[x][1]]=x;
father[tree[x][1]]=0;
tree[x][1]=0;
while (pp[x]){
y=pp[x];
splay(y,0);
//clear(y);
if (tree[y][1]) pp[tree[y][1]]=y;
father[tree[y][1]]=0;
tree[y][1]=x;
father[x]=y;
pp[x]=0;
splay(x,0);
}
}
void link(ll x,ll y){
splay(x,0);
access(y);
splay(y,0);
size[y]+=size[x];
add[y]+=size[x];
pp[x]=y;
}
void cut(ll x,ll y){
splay(x,0);
access(y);
splay(y,0);
size[y]-=size[x];
add[y]-=size[x];
pp[x]=0;
}
void insert(char ch){
ll np=++tot;
size[np]=1;
step[np]=step[last]+1;
ll p=last;
while (p&&g[p][ch-'A']==0){
g[p][ch-'A']=np;
p=pre[p];
}
if (!p){
pre[np]=1;
link(np,1);
}
else{
ll q=g[p][ch-'A'];
if (step[q]==step[p]+1){
pre[np]=q;
link(np,q);
}
else{
ll nq=++tot;
step[nq]=step[p]+1;
pre[nq]=pre[q];
link(nq,pre[q]);
cut(q,pre[q]);
pre[q]=nq;
link(q,nq);
int i;
fo(i,0,25) g[nq][i]=g[q][i];
pre[np]=nq;
link(np,nq);
while (p&&g[p][ch-'A']==q){
g[p][ch-'A']=nq;
p=pre[p];
}
}
}
last=np;
}
int get(){
int x=1,i;
fo(i,1,n){
if (!x) return 0;
x=g[x][s[i]-'A'];
}
if (!x) return 0;
splay(x,0);
return size[x];
}
char get_c(){
char ch=getchar();
while (ch<'A'||ch>'Z') ch=getchar();
return ch;
}
int main(){
//freopen("2555.in","r",stdin);freopen("2555.out","w",stdout);
scanf("%lld",&m);
pre[last=tot=1]=0;
scanf("%s",s+1);
n=strlen(s+1);
fo(i,1,n) insert(s[i]);
mask=0;
while (m--){
ch=get_c();
if (ch=='A') czy=1;else czy=0;
fo(i,1,4-2*czy) ch=get_c();
get_s(mask);
if (czy){
fo(i,1,n) insert(s[i]);
}
else{
ans=get();
mask^=ans;
printf("%lld\n",ans);
}
}
//fclose(stdin);fclose(stdout);
return 0;
}