「SCOI2014」方伯伯的 OJ

和列队有点像，平衡树点分裂维护即可

但是需要额外用个set之类的对编号查找点的位置

插入完了后记得splay，删除时注意特判好多东西

Code:

#include <cstdio>
#include <cctype>
#include <set>
const int N=2e5+10;
template <class T>
void inline read(T &x)
{
    x=0;char c=getchar();
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) x=x*10+c-'0',c=getchar();
}
#define ls ch[now][0]
#define rs ch[now][1]
#define fa par[now]
int siz[N],ch[N][2],L[N],R[N],par[N],tot,root;
void connect(int f,int now,int typ){ch[fa=f][typ]=now;}
int identity(int now){return ch[fa][1]==now;}
void updata(int now){siz[now]=siz[ls]+siz[rs]+R[now]+1-L[now];}
void Rotate(int now)
{
    int p=fa,typ=identity(now);
    connect(p,ch[now][typ^1],typ);
    connect(par[p],now,identity(p));
    connect(now,p,typ^1);
    updata(p),updata(now);
}
void splay(int now,int to)
{
    to=par[to];
    if(!to) root=now;
    for(;fa!=to;Rotate(now))
        if(par[fa]!=to)
            Rotate(identity(now)^identity(fa)?now:fa);
}
struct yuucute
{
    int x,id;
    yuucute(){}
    yuucute(int X,int Id){x=X,id=Id;}
    bool friend operator <(yuucute a,yuucute b){return a.x<b.x;}
    bool friend operator ==(yuucute a,yuucute b){return a.x==b.x;}
};
std::set <yuucute> s;
std::set <yuucute>::iterator it;
#define yuulovely 1
int New(int l,int r)
{
    siz[++tot]=r+1-l,L[tot]=l,R[tot]=r;
    return tot;
}
int getnum(int x)
{
    it=--s.upper_bound(yuucute(x,yuulovely));
    return it->id;
}
void split(int now,int x)
{
    if(L[now]==R[now]) return;
    s.erase(yuucute(L[now],yuulovely));
    s.insert(yuucute(x,now));
    int lson=ls,rson=rs;
    if(L[now]<x)
    {
        int lp=New(L[now],x-1);
        s.insert(yuucute(L[now],lp));
        connect(now,lp,0);
        connect(lp,lson,0);
        updata(lp);
    }
    if(x<R[now])
    {
        int rp=New(x+1,R[now]);
        s.insert(yuucute(x+1,rp));
        connect(now,rp,1);
        connect(rp,rson,1);
        updata(rp);
    }
    L[now]=R[now]=x;
}
void insl(int now,int ins)
{
    ++siz[now];
    if(ls) insl(ls,ins);
    else connect(now,ins,0);
}
void insr(int now,int ins)
{
    ++siz[now];
    if(rs) insr(rs,ins);
    else connect(now,ins,1);
}
int getlef(int now)
{
    if(ls) return getlef(ls);
    return now;
}
void erase(int now)
{
    if(!rs) {par[root=ls]=0,ls=0,updata(now);return;}
    splay(root=getlef(rs),rs);
    connect(root,ls,0);
    updata(root),par[root]=0;
    ls=rs=0,updata(now);
}
int getrank(int now,int &x)
{
    if(siz[ls]>=x) return getrank(ls,x);
    x-=siz[ls];
    if(x<=R[now]-L[now]+1) return now;
    x-=R[now]-L[now]+1;
    return getrank(rs,x);
}
int main()
{
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    int n,m;
    read(n),read(m);
    root=New(1,n);
    s.insert(yuucute(1,root));
    int op,x,y,las=0;
    for(int i=1;i<=m;i++)
    {
        read(op),read(x);
        x-=las;
        if(op==1)
        {
            read(y),y-=las;
            int now=getnum(x);
            splay(now,root);
            printf("%d\n",las=siz[ls]+x+1-L[now]);
            split(now,x);
            L[now]=R[now]=y;
            s.erase(yuucute(x,yuulovely));
            s.insert(yuucute(y,now));
        }
        else if(op==2)
        {
            int now=getnum(x);
            splay(now,root);
            printf("%d\n",las=siz[ls]+x+1-L[now]);
            split(now,x);
            erase(now);
            insl(root,now);
            splay(now,root);
        }
        else if(op==3)
        {
            int now=getnum(x);
            splay(now,root);
            printf("%d\n",las=siz[ls]+x+1-L[now]);
            split(now,x);
            erase(now);
            insr(root,now);
            splay(now,root);
        }
        else
        {
            int now=getrank(root,x);
            printf("%d\n",las=x+L[now]-1);
            splay(now,root);
        }
    }
    return 0;
}

2019.2.23

标签：SCOI2014,OJ,int,void,ch,fa,解题,yuucute,now
来源： https://www.cnblogs.com/butterflydew/p/10421797.html