【BZOJ 3196】二逼平衡树 线段树套splay 模板题

我写的是线段树套splay，网上很多人写的都是套treap，然而本蒟蒻并不会treap

奉上sth神犇的模板：

//bzoj3196 二逼平衡树，支持修改某个点的值，查询区间第k小值，查询区间某个值排名，查询区间某个值值前驱、后继。查询第k小值是log^3(n)的，其他都是log^2(n)的
#include <cstdio>
using namespace std;
const int maxn=2000000,inf=999999999;
int a[maxn],f[maxn],sum[maxn],num[maxn],n,m,c[50001],root[200000],left,right,tot,son[maxn][2];
int mmin(int x,int y)
{
    if (x<y) return x;
    return y;
}
int mmax(int x,int y)
{
    if (x>y) return x;
    return y;
}
void rotate(int x,int w)
{
    int y=f[x];
    if (f[y]) if (y==son[f[y]][0]) son[f[y]][0]=x; else son[f[y]][1]=x;
    f[x]=f[y];
    if (son[x][w]) f[son[x][w]]=y;
    son[y][1-w]=son[x][w];
    f[y]=x;
    son[x][w]=y;
    sum[y]=sum[son[y][0]]+sum[son[y][1]]+num[y];
}
void splay(int r,int x,int w)
{
    int y;
    while (f[x]!=w)
    {
        y=f[x];
        if (f[y]==w) if (x==son[y][0]) rotate(x,1); else rotate(x,0);
        else if (y==son[f[y]][0]) if (x==son[y][0]) {rotate(y,1); rotate(x,1); } else {rotate(x,0); rotate(x,1);}
                else if (x==son[y][0]) {rotate(x,1); rotate(x,0);} else {rotate(y,0); rotate(x,0);}
    }
    sum[x]=sum[son[x][0]]+sum[son[x][1]]+num[x];
    if (w==0) root[r]=x;
}
void insert(int r,int value)
{
    int x=root[r];
    if (x==0)
    {
        a[++tot]=value;
        num[tot]=sum[tot]=1;
        root[r]=tot;
        return;
    }
    while (a[x]!=value)
        if (a[x]>value) {if (son[x][0]) x=son[x][0]; else break;}
        else {if (son[x][1]) x=son[x][1]; else break;}
    if (a[x]==value) {num[x]++; splay(r,x,0); return;}
    a[++tot]=value;
    f[tot]=x;
    num[tot]=sum[tot]=1;
    if (value<a[x]) son[x][0]=tot; else son[x][1]=tot;
    splay(r,tot,0);
}
void del(int r,int value)
{
    int x=root[r];
    while (a[x]!=value)
        if (a[x]>value) {if (son[x][0]) x=son[x][0]; else break;}
        else {if (son[x][1]) x=son[x][1]; else break;}
    splay(r,x,0);
    if (num[x]>1) {num[x]--; return;}
    if (!son[x][0]) {root[r]=son[x][1]; f[son[x][1]]=0; return;}
    if (!son[x][1]) {root[r]=son[x][0]; f[son[x][0]]=0; return;}
    int y=son[x][0];
    while (son[y][1]) y=son[y][1];
    splay(r,y,x);
    son[y][1]=son[x][1];
    f[y]=0;//一定记得f[y]=0
    f[son[y][1]]=y;
    sum[y]=sum[son[y][0]]+sum[son[y][1]]+num[y];
    root[r]=y;
}
void build(int now,int l,int r)
{
    //printf("now=%d l=%d r=%d\n",now,l,r);
    int i;
    for (i=l;i<=r;i++) insert(now,c[i]);//对每个线段树区间都建立一棵平衡树
    //print(now);
    if (l==r) return;
    int mid=(l+r)>>1;
    i=now<<1;
    build(i,l,mid);
    build(i+1,mid+1,r);
}
int find(int r,int value)
{
    int x=root[r];
    while (a[x]!=value)
        if (a[x]>value) {if (son[x][0]) x=son[x][0]; else break;}
        else {if (son[x][1]) x=son[x][1]; else break;}
    splay(r,x,0);
    return x;
}
int rank(int r,int k)
{
    int x=find(r,k);
    if (a[x]>=k) return (sum[son[x][0]]);
    return (sum[son[x][0]]+num[x]);
}
int getrank(int now,int l,int r,int k)
{
    if (l>=left&&r<=right) return (rank(now,k));
    int mid=(l+r)>>1,w=now<<1,ret=0;
    if (left<=mid) ret=getrank(w,l,mid,k);
    if (right>mid) ret+=getrank(w+1,mid+1,r,k);
    return ret;
}
void change(int now,int l,int r,int pos,int value)
{
    del(now,c[pos]);
    insert(now,value);
    if (l==r) return;
    int mid=(l+r)>>1,w=now<<1;
    if (pos<=mid) change(w,l,mid,pos,value); else change(w+1,mid+1,r,pos,value);
}
int ppre(int r,int value)
{
    int x=root[r];
    while (a[x]!=value)
        if (a[x]>value) {if (son[x][0]) x=son[x][0]; else break;}
        else {if (son[x][1]) x=son[x][1]; else break;}
    if (a[x]==value)
    {
        splay(r,x,0);
        if (!son[x][0]) return 1;
        x=son[x][0];
        while (son[x][1]) x=son[x][1];
        return x;
    }
    while (f[x]&&a[x]>value) x=f[x];
    if (a[x]<value) return x;
    return 1;
}
int ssucc(int r,int value)
{
    int x=root[r];
    while (a[x]!=value)
        if (a[x]>value) {if (son[x][0]) x=son[x][0]; else break;}
        else {if (son[x][1]) x=son[x][1]; else break;}
    if (a[x]==value)
    {
        splay(r,x,0);
        if (!son[x][1]) return 2;
        x=son[x][1];
        while (son[x][0]) x=son[x][0];
        return x;
    }
    while (f[x]&&a[x]<value) x=f[x];
    if (a[x]>value) return x;
    return 2;
}
int pre(int now,int l,int r,int value)
{
    if (l>=left&&r<=right)
    {
        int x=ppre(now,value);
        if (a[x]<value) return a[x];
        x=find(now,4);
        return -1;
    }
    int mid=(l+r)>>1,w=now<<1,ret=-1;
    if (left<=mid) ret=mmax(ret,pre(w,l,mid,value));
    if (right>mid) ret=mmax(ret,pre(w+1,mid+1,r,value));
    return ret;
}
int succ(int now,int l,int r,int value)
{
    if (l>=left&&r<=right)
    {
        int x=ssucc(now,value);
        if (a[x]>value) return (a[x]);
        return inf;
    }
    int mid=(l+r)>>1,w=now<<1,ret=inf;
    if (left<=mid) ret=mmin(ret,succ(w,l,mid,value));
    if (right>mid) ret=mmin(ret,succ(w+1,mid+1,r,value));
    return ret;
}
int main()
{
    scanf("%d%d",&n,&m);
    int i,kind,x,y,z,min,max,ans,mid;
    for (i=1;i<=n;i++) scanf("%d",&c[i]);
    tot=2;
    a[1]=-1,a[2]=999999999;
    build(1,1,n);
    for (i=1;i<=m;i++)
    {
        scanf("%d%d%d",&kind,&x,&y);
        if (kind!=3) scanf("%d",&z);
        if (kind==1)
        {
            left=x,right=y;
            printf("%d\n",getrank(1,1,n,z)+1);
            continue;
        }
        if (kind==2)
        {
            left=x,right=y;
            min=0,max=100000000;
            while (min<=max)
            {
                mid=(min+max)>>1;
                if (getrank(1,1,n,mid)<z) {ans=mid; min=mid+1;} else max=mid-1;
            }
            printf("%d\n",ans);
            continue;
        }
        if (kind==3)
        {
            change(1,1,n,x,y);
            c[x]=y;
            continue;
        }
        if (kind==4)
        {
            left=x,right=y;
            printf("%d\n",pre(1,1,n,z));
            continue;
        }
        left=x,right=y;
        printf("%d\n",succ(1,1,n,z));
    }
    return 0;
}

View Code

然后是我的AC code:

/*调了好久，一部分原因是做这道题时学长讲新课，没时间调这道题。今天终于调出来了，然后把三篇平衡树的博客一起写出来了。这个题的题解在上面sth神犇的模板里有，主要是那个二分k值是本题的特色。我这道题一直WA的原因是快速读入没有判断负号(╯‵□′)╯︵┻━┻*/

#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
using namespace std;
inline const int max(const int &a,const int &b){return a>b?a:b;}
inline const int min(const int &a,const int &b){return a<b?a:b;}
struct node{
    node();
    node *fa,*ch[2];
    int sum,d;
    short pl(){return this==fa->ch[1];}
    void count(){sum=ch[0]->sum+ch[1]->sum+1;}
}*null;
node::node(){fa=ch[1]=ch[0]=null;sum=0;}
int data[50003],N,M;
node *ROOT[50000<<2];
int getint(){char c;int fh=1;while(!isdigit(c=getchar()))if (c=='-')fh=-1;int a=c-'0';while(isdigit(c=getchar()))a=a*10+c-'0';return a*fh;}
namespace Splay{
    void Builda(){
        null=new node;
        *null=node();
    }
    void rotate(node *k,int rt){
        node *r=k->fa;
        if (k==null||r==null) return;
        int x=k->pl()^1;
        r->ch[x^1]=k->ch[x];
        r->ch[x^1]->fa=r;
        if (r->fa==null) ROOT[rt]=k;
        else r->fa->ch[r->pl()]=k;
        k->fa=r->fa; r->fa=k;
        k->ch[x]=r;
        r->count(); k->count();
    }
    void splay(int rt,node *r,node *tar=null){
        for (;r->fa!=tar;rotate(r,rt))
         if (r->fa->fa!=tar)rotate(r->pl()==r->fa->pl()?r->fa:r,rt);
    }
    void insert(int rt,int x){
        node *r=ROOT[rt];
        if (ROOT[rt]==null){
            ROOT[rt]=new node;
            *ROOT[rt]=node();
            ROOT[rt]->d=x;
            ROOT[rt]->count();
            return;
        }
        while (1){
            int c;
            if (x<r->d) c=0;
            else c=1;
            if (r->ch[c]==null){
                r->ch[c]=new node;
                *r->ch[c]=node();
                r->ch[c]->fa=r;
                r->ch[c]->d=x;
                splay(rt,r->ch[c]);
                return;
            }else r=r->ch[c];
        }
    }
    void build(int rt,int l,int r){
        ROOT[rt]=null;
        for(int i=l;i<=r;++i) insert(rt,data[i]);
    }
    node *kth(int rt,int x){
        node *r=ROOT[rt];
        while (r!=null){
            if (x<r->d) r=r->ch[0];
            else if (x>r->d) r=r->ch[1];
            else return r;
        }return r;
    }
    node *rightdown(node *r){
        while (r->ch[1]!=null){
            r=r->ch[1];
        }return r;
    }
    void deletee(int rt,int x){
        node *r=kth(rt,x);
        splay(rt,r);
        if ((r->ch[0]==null)&&(r->ch[1]==null)){
            ROOT[rt]=null;
            delete r;
        }else if (r->ch[0]==null){
            r->ch[1]->fa=null;
            ROOT[rt]=r->ch[1];
            delete r;
        }else if (r->ch[1]==null){
            r->ch[0]->fa=null;
            ROOT[rt]=r->ch[0];
            delete r;
        }else{
            splay(rt,rightdown(r->ch[0]),ROOT[rt]);
            r->ch[0]->ch[1]=r->ch[1];
            r->ch[1]->fa=r->ch[0];
            r->ch[0]->fa=null;
            r->ch[0]->count();
            ROOT[rt]=r->ch[0];
            delete r;
        }
    }
    int predd(node *r,int x){
        if (r==null) return -1;
        if (x<=r->d) return predd(r->ch[0],x);
        else return max(r->d,predd(r->ch[1],x));
    }
    int pross(node *r,int x){
        if (r==null) return 1E8+10;
        if (x>=r->d) return pross(r->ch[1],x);
        else return min(r->d,pross(r->ch[0],x));
    }
    node *get1(node *r,int x){
        if (r==null) return null;
        if (x<r->d) return get1(r->ch[0],x);
        if (x>r->d) return get1(r->ch[1],x);
        node *rr=get1(r->ch[0],x);
        if (rr!=null) return rr; else return r;
    }
    int quee1(int rt,int x){
        node *r=get1(ROOT[rt],x);
        if (r!=null){
            splay(rt,r);
            return r->ch[0]->sum;
        }else{
            insert(rt,x);
            int anss=ROOT[rt]->ch[0]->sum;
            deletee(rt,x);
            return anss;
        }
    }
    int quee2(int rt,int x){
        node *r=get1(ROOT[rt],x);
        if (r!=null){
            splay(rt,r);
            return r->ch[0]->sum;
        }else{
            insert(rt,x);
            int anss=ROOT[rt]->ch[0]->sum;
            deletee(rt,x);
            return anss;
        }
    }
}
void buildtree(int l,int r,int rt){
    Splay::build(rt,l,r);
    if (l==r) {Splay::build(rt,l,r);return;}
    int mid=(l+r)>>1;
    buildtree(l,mid,rt<<1);
    buildtree(mid+1,r,rt<<1|1);
}
int que1(int L,int R,int k,int l,int r,int rt){
    if ((L<=l)&&(r<=R)) return Splay::quee1(rt,k);
    int mid=(l+r)>>1,s=0;
    if (L<=mid) s+=que1(L,R,k,l,mid,rt<<1);
    if (R>mid) s+=que1(L,R,k,mid+1,r,rt<<1|1);
    return s;
}
int que2(int L,int R,int k,int l,int r,int rt){
    if ((L<=l)&&(r<=R)) return Splay::quee2(rt,k);
    int mid=(l+r)>>1,s=0;
    if (L<=mid) s+=que2(L,R,k,l,mid,rt<<1);
    if (R>mid) s+=que2(L,R,k,mid+1,r,rt<<1|1);
    return s;
}
void que3(int pos,int k,int l,int r,int rt){
    if ((l<=pos)&&(pos<=r)){
        Splay::deletee(rt,data[pos]);
        Splay::insert(rt,k);
    }if (l==r) return; int mid=(l+r)>>1;
    if (pos<=mid) que3(pos,k,l,mid,rt<<1);
    if (pos>mid) que3(pos,k,mid+1,r,rt<<1|1);
}
int que4(int L,int R,int k,int l,int r,int rt){
    if ((L<=l)&&(r<=R)) return Splay::predd(ROOT[rt],k);
    int mid=(l+r)>>1,s=-1;
    if (L<=mid) s=que4(L,R,k,l,mid,rt<<1);
    if (R>mid) s=max(s,que4(L,R,k,mid+1,r,rt<<1|1));
    return s;
}
int que5(int L,int R,int k,int l,int r,int rt){
    if ((L<=l)&&(r<=R)) return Splay::pross(ROOT[rt],k);
    int mid=(l+r)>>1,s=1E8+10;
    if (L<=mid) s=que5(L,R,k,l,mid,rt<<1);
    if (R>mid) s=min(s,que5(L,R,k,mid+1,r,rt<<1|1));
    return s;
}
int main(){
    Splay::Builda();
    N=getint();M=getint();
    for(int i=1;i<=N;++i)data[i]=getint();
    buildtree(1,N,1);
    while (M){M--;
        int x=getint();
        switch (x){
            int l,r,k,pos,ans,left,right,mid;
            case 1:
                l=getint(),r=getint(),k=getint();
                printf("%d\n",que1(l,r,k,1,N,1)+1);
            break;
            case 2:
                l=getint(),r=getint(),k=getint(),left=0,right=1E8;
                while (left<=right){
                    mid=(left+right)>>1;
                    if (que2(l,r,mid,1,N,1)+1<=k) ans=mid,left=mid+1;
                    else right=mid-1;
                }printf("%d\n",ans);
            break;
            case 3:
                pos=getint(),k=getint();
                que3(pos,k,1,N,1);
                data[pos]=k;
            break;
            case 4:
                l=getint(),r=getint(),k=getint();
                printf("%d\n",que4(l,r,k,1,N,1));
            break;
            case 5:
                l=getint(),r=getint(),k=getint();
                printf("%d\n",que5(l,r,k,1,N,1));
            break;
        }
    }
    return 0;
}

这样就可以了