BZOJ3065：带插入区间K小值

浅谈树状数组与主席树：https://www.cnblogs.com/AKMer/p/9946944.html

题目传送门：https://www.lydsy.com/JudgeOnline/problem.php?id=3065

去%了一发hzwer的博客学会的。

替罪羊树套线段树，外层维护位置，内层维护权值。

时间复杂度：\(O(nlog^2n)\)

空间复杂度：\(O(nlog^2n)\)

代码如下：

#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
 
const double alpha=0.75;
const int maxn=7e4+5,maxsz=2e7;
 
char s[5];
int n,m,lstans;
 
int read() {
    int x=0,f=1;char ch=getchar();
    for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
    for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
    return x*f;
}
 
struct Segment_tree {
    int tot;
    vector<int>rub;
    int sum[maxsz],ls[maxsz],rs[maxsz];
 
    int newnode() {
        if(rub.empty())return ++tot;
        else {
            int res=rub.back();rub.pop_back();
            return res;
        }
    }
 
    void change(int p,int l,int r,int pos,int v) {
        while(1) {
            sum[p]+=v;
            if(l==r)return;
            int mid=(l+r)>>1;
            if(pos<=mid) {
                if(!ls[p])ls[p]=newnode();
                p=ls[p],r=mid;
            }
            else {
                if(!rs[p])rs[p]=newnode();
                p=rs[p],l=mid+1;
            }
        }
    }
 
    void recycle(int u) {
        if(!u)return;
        rub.push_back(u);
        recycle(ls[u]),recycle(rs[u]);
        sum[u]=ls[u]=rs[u]=0;
    }
}T2;
 
struct TiZuiYang_tree { 
    vector<int>num;
    int tot,root,cnt;
    int son[maxn][2];
    vector<int>::iterator it;
    int rt[maxn],siz[maxn],tmp[maxn],a[maxn],id[maxn];
 
    int build(int l,int r) {
        if(r<l)return 0;
        if(l==r) {
            int u=id[l];son[u][0]=son[u][1]=0;
            rt[u]=T2.newnode(),siz[u]=1;
            T2.change(rt[u],0,70000,a[u],1);
            return u;
        }
        int mid=(l+r)>>1,u=id[mid];
        rt[u]=T2.newnode();siz[u]=1;
        son[u][0]=build(l,mid-1);
        son[u][1]=build(mid+1,r);
        siz[u]+=siz[son[u][0]]+siz[son[u][1]];
        for(int i=l;i<=r;i++)T2.change(rt[u],0,70000,a[id[i]],1);
        return u;
    }
 
    void find(int u,int l,int r) {
        if(l==1&&r==siz[u]) {tmp[++cnt]=rt[u];return;}
        int sz=siz[son[u][0]];
        if(l<=sz+1&&r>=sz+1)num.push_back(a[u]);
        if(r<=sz)find(son[u][0],l,r);
        else if(l>sz+1)find(son[u][1],l-sz-1,r-sz-1);
        else {
            if(l<=sz)find(son[u][0],l,sz);
            if(r>sz+1)find(son[u][1],1,r-sz-1);
        }
    }
 
    int query(int l,int r,int rk) {
        num.clear(),cnt=0,find(root,l,r);
        int L=0,R=70000;
        while(L!=R) {
            int res=0,mid=(L+R)>>1;
            for(int i=1;i<=cnt;i++)
                res+=T2.sum[T2.ls[tmp[i]]];
            for(it=num.begin();it!=num.end();it++)
                if(*it>=L&&(*it)<=mid)res++;
            if(res>=rk) {
                for(int i=1;i<=cnt;i++)
                    tmp[i]=T2.ls[tmp[i]];
                R=mid;
            }
            else {
                for(int i=1;i<=cnt;i++)
                    tmp[i]=T2.rs[tmp[i]];
                L=mid+1;rk-=res;
            }
        }
        return L;
    }
 
    int change(int u,int pos,int x) {
        T2.change(rt[u],0,70000,x,1);
        int sz=siz[son[u][0]],res;
        if(sz+1==pos) {res=a[u];a[u]=x;}
        else if(sz>=pos)res=change(son[u][0],pos,x);
        else res=change(son[u][1],pos-sz-1,x);
        T2.change(rt[u],0,70000,res,-1);
        return res;
    }
 
    void insert(int &u,int rk,int v) {
        if(!u) {
            u=++tot;rt[u]=T2.newnode();
            T2.change(rt[u],0,70000,v,1);
            siz[u]=1;a[u]=v;return;
        }
        T2.change(rt[u],0,70000,v,1);
        int sz=siz[son[u][0]];siz[u]++;
        if(sz>=rk)insert(son[u][0],rk,v);
        else insert(son[u][1],rk-sz-1,v);
    }
 
    void dfs_rebuild(int u) {
        if(!u)return;
        dfs_rebuild(son[u][0]);
        T2.recycle(rt[u]);
        id[++cnt]=u;rt[u]=0;
        dfs_rebuild(son[u][1]);
    }
 
    void find_rebuild(int &u,int rk) {
        if(alpha*siz[u]<max(siz[son[u][0]],siz[son[u][1]])) {
            cnt=0;dfs_rebuild(u);u=build(1,cnt);
            return;
        }
        if(siz[son[u][0]]>=rk)find_rebuild(son[u][0],rk);
        if(rk>siz[son[u][0]]+1)find_rebuild(son[u][1],rk-siz[son[u][0]]-1);
    }
}T1;
 
int main() {
    T1.tot=n=read();
    for(int i=1;i<=n;i++)
        T1.id[i]=i,T1.a[i]=read();
    T1.root=T1.build(1,n);
    m=read();
    for(int i=1;i<=m;i++) {
        scanf("%s",s+1);
        if(s[1]=='Q') {
            int l=read()^lstans,r=read()^lstans,k=read()^lstans;
            lstans=T1.query(l,r,k);printf("%d\n",lstans);
        }
        if(s[1]=='M') {
            int pos=read()^lstans,x=read()^lstans;
            T1.change(T1.root,pos,x);
        }
        if(s[1]=='I') {
            int pos=read()^lstans,x=read()^lstans;
            T1.insert(T1.root,pos-1,x);
            T1.find_rebuild(T1.root,pos);
        }
    }
    return 0;
}