和弹飞绵羊类似的,那题是维护一个内向树,这题是维护若干棵基环内向树和一棵内向树
同样的用LCT维护就好了
内向树可以直接维护,对于基环内向树,随便找环上一点x做根,记录他指向哪个点to[x]
对于Cut操作,不在环上可以直接断,否则断了之后要把x和to[x]连起来
对于Link操作,如果不成环同样可以直接连,否则直接把这个Link的点当作这棵基环内向树的根就行了
询问的话,如果不在n+1那棵内向树上直接输出-1,否则答案就是他的层数
code:
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<complex>
#include<iostream>
#include<algorithm>
#define ll long long
#define pb push_back
using namespace std;
inline void read(int &x)
{
char c; int f=1;
while(!((c=getchar())>='0'&&c<='9')) if(c=='-') f=-1;
x=c-'0';
while((c=getchar())>='0'&&c<='9') (x*=10)+=c-'0';
if(f==-1) x=-x;
}
const int maxn = 210000;
int n,m;
int a[maxn];
struct Link_Cut_Tree
{
int son[maxn][2],fa[maxn],siz[maxn];
void init()
{
siz[0]=0;
for(int i=1;i<=n+1;i++) siz[i]=1;
}
void pushup(const int x){ siz[x]=siz[son[x][0]]+siz[son[x][1]]+1; }
bool isrt(const int x){ return son[fa[x]][0]!=x&&son[fa[x]][1]!=x; }
void rot(int x)
{
int y=fa[x],z=fa[y];
if(!isrt(y)) son[z][son[z][1]==y]=x;
fa[x]=z;
int l=son[y][1]==x;
fa[son[y][l]=son[x][!l]]=y;
fa[son[x][!l]=y]=x;
pushup(y);
}
void splay(int x)
{
for(;!isrt(x);rot(x))
{
int y=fa[x],z=fa[y];
if(!isrt(y)) rot(((son[z][1]==y)^(son[y][1]==x))?x:y);
}pushup(x);
}
void Access(int x)
{
for(int y=0;x;y=x,x=fa[x])
{
splay(x); son[x][1]=y;
pushup(x);
}
}
int go(int x,int dir)
{
while(son[x][dir]) x=son[x][dir];
return x;
}
int findrt(int x)
{
Access(x); splay(x);
return go(x,0);
}
bool Connected(int x,int y){ return findrt(x)==findrt(y); }
void Link(int x,int y)
{
splay(y); fa[y]=x;
}
void Cut(int x)
{
Access(x); splay(x);
fa[son[x][0]]=0,son[x][0]=0,pushup(x);
}
int q(int x)
{
Access(x); splay(x);
return siz[son[x][0]];
}
bool InCir(int x)
{
int ff=findrt(x);
Access(a[ff]);
splay(x);
return go(x,0)==ff;
}
}LCT;
int main()
{
read(n); read(m);
for(int i=1;i<=n;i++)
{
int x; read(x);
if(i+x>n||i+x<=0) a[i]=n+1;
else a[i]=i+x;
}
LCT.init();
for(int i=1;i<=n;i++)
{
int y=i,x=a[i];
if(LCT.Connected(x,y)) continue;
else LCT.Link(x,y);
}
for(int i=1;i<=m;i++)
{
int op; read(op);
if(op==1)
{
int x;read(x);
if(!LCT.Connected(x,n+1)) puts("-1");
else printf("%d\n",LCT.q(x));
}
else
{
int x,c; read(x); read(c);
c=(x+c>n||x+c<=0)?n+1:x+c;
if(LCT.Connected(x,n+1)||!LCT.InCir(x)) LCT.Cut(x);
else if(LCT.findrt(x)!=x)
{
int ff=LCT.findrt(x);
LCT.Cut(x);
LCT.Link(a[ff],ff);
}
if(LCT.findrt(c)!=x) LCT.Link(c,x);
a[x]=c;
}
}
return 0;
}