BZOJ传送门

洛谷传送门


解析:

LCTLCT裸题啊。。。

思路:

可以很显然的发现不管怎么变,我们设置虚拟节点n+1n+1,所有点到它的路径构成一棵树。

那不就完了,直接LCTLCT维护这棵路径树的形态,路径上经过多少点就是要被弹多少次,在LCTLCT的节点里维护sizsiz,最后提取路径,siz1siz-1就是最终答案。


代码:

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define re register
#define gc getchar
#define pc putchar
#define cs const

inline
int getint(){
	re int num;
	re char c;
	while(!isdigit(c=gc()));num=c^48;
	while(isdigit(c=gc()))num=(num<<1)+(num<<3)+(c^48);
	return num;
}

inline
void outint(int a){
	static char ch[13];
	if(a==0)pc('0');
	while(a)ch[++ch[0]]=a-a/10*10,a/=10;
	while(ch[0])pc(ch[ch[0]--]^48);
}

typedef struct splay_node *point;
struct splay_node{
	point son[2],fa;
	int siz;
	bool tag;
	
	point &lc(){return son[0];}
	point &rc(){return son[1];}
	
	bool which(){return fa->son[1]==this;}
	bool isroot(){return !fa||(fa->son[0]!=this&&fa->son[1]!=this);}
	
	void pushdown(){
		if(tag){
			swap(son[0],son[1]);
			if(son[0])son[0]->tag^=1;
			if(son[1])son[1]->tag^=1;
			tag=0; 
		}
	}
	void pushup(){
		siz=(son[0]?son[0]->siz:0)+(son[1]?son[1]->siz:0)+1;
	}
	
};

cs int N=200005;
struct Link_Cut_Tree{
	splay_node ttt[N];
	
	void Rotate(point now){
		point Fa=now->fa,FA=Fa->fa;
		bool pos=now->which();
		if(FA&&!Fa->isroot())FA->son[Fa->which()]=now;
		now->fa=FA;
		Fa->fa=now;
		Fa->son[pos]=now->son[!pos];
		if(Fa->son[pos])Fa->son[pos]->fa=Fa;
		now->son[!pos]=Fa;
		Fa->pushup();
		now->pushup();
	}
	
	void Splay(point now){
		static point q[N];
		static int qn;
		q[qn=1]=now;
		for(point Fa=now;!Fa->isroot();Fa=Fa->fa)q[++qn]=Fa->fa;
		for(int re i=qn;i;--i)q[i]->pushdown();
		for(point Fa;!now->isroot();Rotate(now))
		if(!(Fa=now->fa)->isroot())Rotate(now->which()==Fa->which()?Fa:now);
	}
	
	void access(point now){
		for(point son=NULL;now;son=now,now=now->fa)
		Splay(now),now->rc()=son,now->pushup();
	}
	
	void makeroot(point now){
		access(now),Splay(now),now->tag^=1;
	}
	
	void link(point u,point v){
		makeroot(u),u->fa=v;
	}
	void link(int u,int v){link(&ttt[u],&ttt[v]);}
	
	void cut(point u,point v){
		makeroot(u),access(v),Splay(v);
		v->lc()=u->fa=NULL;
	}
	void cut(int u,int v){cut(&ttt[u],&ttt[v]);}
	
	int query(point u,point v){
		makeroot(u),access(v),Splay(v);
		return v->siz-1;
	}
	int query(int u,int v){return query(&ttt[u],&ttt[v]);}
	
}LCT;

int n,m;
int val[N];
signed main(){
	n=getint();
	for(int re i=1;i<=n;++i){
		val[i]=getint();
		int to=(val[i]+i>n)?(n+1):(val[i]+i);
		LCT.link(i,to);
	}
	m=getint();
	while(m--){
		int op=getint();
		int x=getint()+1;
		switch(op){
			case 1:{
				outint(LCT.query(x,n+1));pc('\n');
				break;
			}
			case 2:{
				int k=getint();
				int to=(val[x]+x>n)?(n+1):(val[x]+x);
				LCT.cut(x,to);
				val[x]=k;to=(val[x]+x>n)?(n+1):(val[x]+x);
				LCT.link(x,to);
				break;
			}
		}
	}
	return 0;
}
posted on 2018-10-08 14:45  zxyoi_dreamer  阅读(66)  评论(0编辑  收藏