●BZOJ 2002 [Hnoi2010]Bounce 弹飞绵羊

Posted 2020-10-22 *ZJ

 

题链：

http://www.lydsy.com/JudgeOnline/problem.php?id=2002

题解：

LCT

如果把弹跳的起点和终点连一条边，弹出去的与n+1号点连边，

则不难发现，整个图形成了一颗树，

同时需要支持树的修改(拆分，合并)和询问点的深度(该点到根的链上的点的个数)，

所以LCT可以很轻松的解决本题。

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
#define MAXN 200050
using namespace std;
int N,M,next[MAXN];
struct LCT{
	int ch[MAXN][2],fa[MAXN],rev[MAXN],size[MAXN];
	bool Which(int x){return ch[fa[x]][1]==x;}
	void Reverse(int x){swap(ch[x][0],ch[x][1]);rev[x]^=1;}
	bool Isroot(int x){return ch[fa[x]][0]!=x&&ch[fa[x]][1]!=x;}
	void Pushup(int x){size[x]=size[ch[x][0]]+size[ch[x][1]]+1;}
	void Pushdown(int x){
		if(!Isroot(x)) Pushdown(fa[x]);
		if(rev[x]) Reverse(ch[x][0]),Reverse(ch[x][1]),rev[x]^=1;
	}
	void Rotate(int x){
		static int y,z,l1,l2;
		y=fa[x]; z=fa[y];
		l1=Which(y); l2=Which(x); fa[x]=z;
		if(!Isroot(y)) ch[z][l1]=x;
		fa[y]=x; fa[ch[x][l2^1]]=y; 
		ch[y][l2]=ch[x][l2^1]; ch[x][l2^1]=y;
		Pushup(y);
	}
	void Splay(int x){
		static int y; Pushdown(x);
		for(;y=fa[x],!Isroot(x);Rotate(x)) if(!Isroot(y)) 
			Rotate(Which(y)==Which(x)?y:x);
		Pushup(x);
	}
	void Access(int x){
		static int y;
		for(y=0;x;y=x,x=fa[x]) 
			Splay(x),ch[x][1]=y,Pushup(x);//!!!
	}
	void Beroot(int x){
		Access(x); Splay(x); Reverse(x);
	}
	void Cut(int x,int y){
		Beroot(x); Access(y); Splay(y);
		fa[x]=ch[y][0]=0; Pushup(y);
	}
	void Link(int x,int y){
		Beroot(x); fa[x]=y;
	}
	void Modify(int x,int k){
		static int t; t=min(x+k,N+1);
		Cut(next[x],x); Link(t,x); next[x]=t;
	}
	int Query(int x){
		Beroot(N+1); Access(x); Splay(x);
		return size[x]-1;
	}
}DT;
int main(){
	scanf("%d",&N); DT.size[N+1]=1;
	for(int i=1,k;i<=N;i++){
		scanf("%d",&k); next[i]=min(i+k,N+1);
		DT.fa[i]=next[i]; DT.size[i]=1;
	} 
	scanf("%d",&M); int a,b,c; 
	for(int i=1;i<=M;i++){
		scanf("%d%d",&a,&b); b++;
		if(a==1) printf("%d\n",DT.Query(b));
		else scanf("%d",&c),DT.Modify(b,c);
	}
	return 0;
}