【洛谷p3391】 Splay模板 文艺平衡树

题目

模板题。

注意标记即可，另外，涉及区间翻转操作，记得设立首尾哨兵。

#include<bits/stdc++.h>
using namespace std;
const int Maxn=0x3f3f3f3f;
const int N=1e5+5;
int lc[N],rc[N],fa[N],sze[N],vi[N],pos[N],a[N];
int n,m,x,y,rt,T;

inline int get(){//快读 
	char ch;bool f=false;int res=0;
	while(((ch=getchar())<'0'||ch>'9')&&ch!='-');
	if(ch=='-') f=true;
	else res=ch-'0';
	while((ch=getchar())>='0'&&ch<='9')
		res=(res<<3)+(res<<1)+ch-'0';
	return f?~res+1:res;
}

inline bool Wrt(const int&x)
{
	return rc[fa[x]]==x;
}

inline void down(const int &x){//下传标记 
	if(x&&pos[x]){
		pos[lc[x]]^=1;
		pos[rc[x]]^=1;
		swap(lc[x],rc[x]);
		pos[x]=0;
	}
}

inline void push(int x){
	sze[x]=sze[lc[x]]+sze[rc[x]]+1;
}

inline int build(int lst,int l,int r){//lst该节点的祖先编号 
	if(l>r) return 0;
	int mid=(l+r)>>1,x=++T;
	fa[x]=lst;pos[x]=0;vi[x]=a[mid];
	lc[x]=build(x,l,mid-1);
	rc[x]=build(x,mid+1,r);
	push(x);
	return x;
}

inline void rot(int x){//单旋 
	int y=fa[x],z=fa[y];
	down(y);down(x);//下传标记 
	int b=(lc[y]==x)?rc[x]:lc[x];
	fa[x]=z;fa[y]=x;
	if(b) fa[b]=y;
	if(z) (y==lc[z]?lc[z]:rc[z])=x;
	if(x==lc[y]) rc[x]=y,lc[y]=b;
	else lc[x]=y,rc[y]=b;
	push(y);push(x);//更新 
}

inline void splay(int x,int tar){
	while(fa[x]!=tar){
		if(fa[fa[x]]!=tar)
			Wrt(x)==Wrt(fa[x])?rot(fa[x]):rot(x);
		rot(x);
	}
	if(!tar) rt=x;
}

inline int Find(int k){//查找第k个数 
	int x=rt,y=k;
	while(x){
		down(x);
		if(y<=sze[lc[x]]) x=lc[x];
		else{
			y-=sze[lc[x]]+1;
			if(!y) return x;
			x=rc[x];
		}
	}
}

/*inline void put(int x){//快输 
	if(x<0){
		x=~x+1;
		putchar('-');
	}
	if(x>9) put(x/10);
	putchar(x%10+48);
}*/

inline void Print(int x){
	down(x);
	if(lc[x]) Print(lc[x]);
	if(vi[x]!=-Maxn&&vi[x]!=Maxn) 
	//	put(vi[x]),putchar(' ');//快输 
	cout<<vi[x]<<" ";
	if(rc[x]) Print(rc[x]);
}

int main(){
	n=get();m=get();
	a[1]=-Maxn;a[n+2]=Maxn;
	for(int i=1;i<=n;i++) a[i+1]=i;
	rt=build(0,1,n+2);
	while(m--){
		int tx=Find(get());
		int ty=Find(get()+2);
		splay(tx,0);
		splay(ty,tx);
		pos[lc[rc[rt]]]^=1;
	} 
	return Print(rt),0;
}

标签：ch,洛谷,lc,fa,int,vi,Splay,p3391,rc
来源： https://blog.csdn.net/C_sat/article/details/123141365