bzoj3065带插入区间K小值

这题其实好像很难，但是听werkeytom_ftd说可以用块链水，于是就很开心地去打了个块状链表套主席树，插入操作就直接插到一个块中，注意如果块的大小2*block就将块分开，注意每一个修改或插入都要修改后继的状态，贴代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>

#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)

using namespace std;

const int N = 35010;
const int M = 610;
const int lim = 70000;

int root[M][M],s[M],a[M][M],next[M],m;
struct point{
    int l,r,tot,p;
}tree[M*M*2*18];
int tot;
int n,q,block,lastans,dep,u1,v1,u2,v2;
int rt[M];
int tt[M*M*2*18],head,tail;

void find(int x,int &u1,int &v1){
    int ss=0,last=0;
    for(int i=1;i;i=next[i]){
        if (ss+s[i]>=x||next[i]==0){
            u1=i;
            v1=x-ss;
            break;
        }
        ss+=s[i];
    }
}

void prepare(){
    fo(i,1,tot)tt[i]=i;
    head=tail=0;
}

int getp(){
    head=head%tot+1;
    return tt[head];
}

void putp(int x){
    tail=tail%tot+1;
    tt[tail]=x;
}

void clear(int x){
    if (!x)return;
    putp(x);
    clear(tree[x].p);
}

void inse(int l,int r,int qf,int &now,int x,int v){
    tree[now=getp()]=tree[qf];
    tree[now].tot+=v;
    if (l==r)return;
    int mid=(l+r)/2;
    if (x<=mid){
        inse(l,mid,tree[qf].l,tree[now].l,x,v);
        tree[now].p=tree[now].l;
    }
    else{
        inse(mid+1,r,tree[qf].r,tree[now].r,x,v);
        tree[now].p=tree[now].r;
    }
}

int getans(int l,int r,int k){
    if (l==r)return l;
    int mid=(l+r)/2,ss=0;
    if (u1!=u2){
        fo(i,v1,s[u1])if (l<=a[u1][i]&&a[u1][i]<=mid)ss++;
        fo(i,1,v2)if (l<=a[u2][i]&&a[u2][i]<=mid)ss++;
        if (next[u1]!=u2)
        for(int i=next[u1];i!=u2;i=next[i])
            ss+=tree[tree[rt[i]].l].tot;
    }
    else fo(i,v1,v2)if (l<=a[u1][i]&&a[u1][i]<=mid)ss++;
    if (ss>=k){
        for(int i=u1;i!=u2;i=next[i])
            rt[i]=tree[rt[i]].l;
        return getans(l,mid,k);
    }
    else{
        for(int i=u1;i!=u2;i=next[i])
            rt[i]=tree[rt[i]].r;
        return getans(mid+1,r,k-ss);
    }
}

int main(){
    scanf("%d",&n);
    tot=360000*2*18;
    prepare();
    block=sqrt(n)+1;
    fo(i,1,n){
        int x;
        scanf("%d",&x);
        int u=(i-1)/block+1,v=(i-1)%block+1;
        inse(0,lim,root[u][v-1],root[u][v],a[u][v]=x,1);
    }
    int u=(n-1)/block;
    fo(i,1,u){
        next[i]=i+1;
        s[i]=block;
    }
    int tim=0;
    s[u+1]=n-u*block;
    m=u+1;
    scanf("%d",&q);
    fo(haha,1,q){
        char ch;
        while(ch=getchar(),ch<‘A‘||ch>‘Z‘);
        if (ch==‘Q‘){
            int x,y,k;
            scanf("%d%d%d",&x,&y,&k);
            x^=lastans;
            y^=lastans;
            k^=lastans;
            find(x,u1,v1);
            find(y,u2,v2);
            for(int i=u1;i!=u2;i=next[i])rt[i]=root[i][s[i]];
            lastans=getans(0,lim,k);
            printf("%d\n",lastans);
            tim++;
        }
        if (ch==‘M‘){
            int x,val;
            scanf("%d%d",&x,&val);
            x^=lastans;
            val^=lastans;
            find(x,u1,v1);
            a[u1][v1]=val;
            fo(i,v1,s[u1]){
                clear(root[u1][i]);
                inse(0,lim,root[u1][i-1],root[u1][i],a[u1][i],1);
            }
        }
        if (ch==‘I‘){
            int x,val;
            scanf("%d%d",&x,&val);
            x^=lastans;
            val^=lastans;
            find(x,u1,v1);
            if (s[u1]+1<block*2){
                s[u1]++;
                fd(i,s[u1],v1+1)a[u1][i]=a[u1][i-1];
                a[u1][v1]=val;
                fo(i,v1,s[u1]){
                    clear(root[u1][i]);
                    inse(0,lim,root[u1][i-1],root[u1][i],a[u1][i],1);
                }
            }
            else{
                if (v1<=block){
                    s[++m]=block;
                    fo(i,1,block)
                    inse(0,lim,root[m][i-1],root[m][i],a[m][i]=a[u1][i+block-1],1);
                    s[u1]=block;
                    next[m]=next[u1];
                    next[u1]=m;
                    fd(i,s[u1],v1+1)a[u1][i]=a[u1][i-1];
                    a[u1][v1]=val;
                    fo(i,v1,s[u1]){
                        clear(root[u1][i]);
                        inse(0,lim,root[u1][i-1],root[u1][i],a[u1][i],1);
                    }
                }
                else{
                    s[++m]=block;
                    v1-=block;
                    fo(i,1,v1-1){
                        a[m][i]=a[u1][i+block];
                        inse(0,lim,root[m][i-1],root[m][i],a[m][i],1);
                    }
                    a[m][v1]=val;
                    inse(0,lim,root[m][v1-1],root[m][v1],val,1);
                    fo(i,v1+1,block){
                        a[m][i]=a[u1][i+block-1];
                        inse(0,lim,root[m][i-1],root[m][i],a[m][i],1);
                    }
                    s[u1]=block;
                    next[m]=next[u1];
                    next[u1]=m;
                }
            }
        }
    }
    return 0;
}

好吧，vfleaking说正解是这样的：http://vfleaking.blog.163.com/blog/static/1748076342013123659818/