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  • 洛谷P5055 【模板】可持久化文艺平衡树(FHQ Treap)

    题面

    传送门

    题解

    日常敲板子.jpg

    //minamoto
    #include<bits/stdc++.h>
    #define R register
    #define inline __inline__ __attribute__((always_inline))
    #define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
    #define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
    #define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
    template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
    template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
    using namespace std;
    char buf[1<<21],*p1=buf,*p2=buf;
    inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
    long long read(){
        R long long res,f=1;R char ch;
        while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
        for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
        return res*f;
    }
    char sr[1<<21],z[20];int K=-1,Z=0;
    inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}
    void print(R long long x){
        if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;
        while(z[++Z]=x%10+48,x/=10);
        while(sr[++K]=z[Z],--Z);sr[++K]='
    ';
    }
    const int N=2e5+5,M=2e7+5;
    typedef long long ll;
    struct node;typedef node* ptr;
    unsigned int aaa=19260817;
    inline unsigned int rd(){aaa^=aaa>>15,aaa+=aaa<<12,aaa^=aaa>>3;return aaa;}
    inline void swap(R ptr &s,R ptr &t){R ptr p=s;s=t,t=p;}
    struct node{
    	ptr lc,rc;bool t;int v,sz;ll sum;unsigned int pr;
    	inline ptr upd(){return sz=lc->sz+rc->sz+1,sum=lc->sum+rc->sum+v,this;}
    	inline ptr init(R int val){return sum=v=val,sz=1,pr=rd(),this;}
    	inline ptr ppd(){return swap(lc,rc),t^=1,this;}
    }e[M],*rt[N],*pp=e;
    inline ptr newnode(R int v){return ++pp,pp->lc=pp->rc=e,pp->init(v);}
    inline ptr cl(ptr p){return ++pp,*pp=*p,pp;}
    inline ptr pd(ptr p){
    	if(p->t){
    		if(p->lc!=e)p->lc=cl(p->lc),p->lc->ppd();
    		if(p->rc!=e)p->rc=cl(p->rc),p->rc->ppd();
    		p->t=0;
    	}return p;
    }
    void split(ptr p,int k,ptr &s,ptr &t){
    	if(p==e)return s=t=e,void();
    	if(pd(p)->lc->sz<k)s=cl(p),split(s->rc,k-s->lc->sz-1,s->rc,t),s->upd();
    		else t=cl(p),split(t->lc,k,s,t->lc),t->upd();
    }
    ptr merge(ptr s,ptr t){
    	if(s==e)return t;if(t==e)return s;
    	if(pd(s)->pr<pd(t)->pr)return s->rc=merge(s->rc,t),s->upd();
    	return t->lc=merge(s,t->lc),t->upd();
    }
    void push(ptr &rt,int k,int x){
    	ptr s,t;
    	split(rt,k,s,t),rt=merge(merge(s,newnode(x)),t);
    }
    void pop(ptr &rt,int k){
    	ptr s,t,p,q;
    	split(rt,k,s,t),split(s,k-1,p,q),rt=merge(p,t);
    }
    void rev(ptr &rt,int l,int r){
    	ptr s,t,p,q;
    	split(rt,r,s,t),split(s,l-1,p,q);
    	rt=merge(merge(p,q->ppd()),t);
    }
    ll query(ptr &rt,int l,int r){
    	ptr s,t,p,q;ll res;
    	split(rt,r,s,t),split(s,l-1,p,q),res=q->sum;
    	return rt=merge(merge(p,q),t),res;
    }
    ll lasans,id,op,l,r,x,k;
    int main(){
    //	freopen("testdata.in","r",stdin);
    	rt[0]=e,e->lc=e->rc=e,lasans=0;
    	for(int Q=read(),i=1;i<=Q;++i){
    		id=read(),op=read(),rt[i]=rt[id];
    		switch(op){
    			case 1:k=read()^lasans,x=read()^lasans,push(rt[i],k,x);break;
    			case 2:k=read()^lasans,pop(rt[i],k);break;
    			case 3:l=read()^lasans,r=read()^lasans,rev(rt[i],l,r);break;
    			case 4:l=read()^lasans,r=read()^lasans,print(lasans=query(rt[i],l,r));break;
    		}
    	}
    	return Ot(),0;
    }
    
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  • 原文地址:https://www.cnblogs.com/bztMinamoto/p/10884044.html
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