[BZOJ 3110] [luogu 3332] [ZJOI 2013] k query large numbers (weights set tree line segment tree)

[BZOJ 3110] [luogu 3332] [ZJOI 2013] k query large numbers (weights set tree line segment tree)

Face questions

Original title face a little ambiguous, but can be seen from the examples really mean

With n positions, each position can be seen as a set.

1 abc: inserting a number c of each set ab

2 abc: 2: Ask each set ab merged together after all the elements of a large c

analysis

The outer layer segment tree maintenance value weights, the inner segment tree with ordinary maintenance position

We first consider the global challenge of the k large, apparently only weights segment tree maintenance occurs case of a global value, interval [L, R] number of elements stored value falls within the [L, R], then as long as the segment tree a-half to

But we have to ask the value range of positions, then the node weights tree line would not only store a value, but rather a storage segment tree (dynamic open points). Is the number of [L, R] of each node of the tree outer segment [L, R], node [l, r] corresponding to an inner segment tree maintenance position, the inner layer maintains a segment tree position [l, r] inner

This section asks the k-big time interval only need to query the tree line in the inner layer.

When we first modified segment tree found in the outer zone comprising c, and the position of the inner segment tree i (a <= i <= b) + 1 requires as much the value of the position on the i. Each of them directly to the inner segment tree to do interval adder

Time complexity \ (O (m \ log ^ 2n) \)

Code

//https://www.luogu.org/problem/P3332 
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 50000
#define maxm 50000
#define maxlogn 60
using namespace std;
typedef long long ll;
inline void qread(int &x) {
    x=0;
    int sign=1;
    char c=getchar();
    while(c<'0'||c>'9') {
        if(c=='-') sign=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9') {
        x=x*10+c-'0';
        c=getchar();
    }
    x=x*sign;
}
inline void qprint(int x) {
    if(x<0) {
        putchar('-');
        qprint(-x);
    } else if(x==0) {
        putchar('0');
        return;
    } else {
        if(x>=10) qprint(x/10);
        putchar('0'+x%10);
    }
}

int n,m;
struct segment_tree_pos{
//内层维护位置的线段树，维护外层权值线段树区间[L,R]内的数的出现情况
//叶子节点存储第i个位置的集合里有多少个在[L,R]内的数，区间求和 
#define lson(x) tree[x].ls
#define rson(x) tree[x].rs
    struct node{
        int ls;
        int rs;
        ll v;
        ll mark;
    }tree[maxn*4*maxlogn+5];
    int ptr=0;
    void push_up(int x){
        tree[x].v=tree[lson(x)].v+tree[rson(x)].v;
    }
    void push_down(int x,int l,int r){
        int mid=(l+r)>>1;
        if(tree[x].mark){
            if(!lson(x)) tree[x].ls=++ptr;
            if(!rson(x)) tree[x].rs=++ptr;
            tree[lson(x)].mark+=tree[x].mark;
            tree[lson(x)].v+=tree[x].mark*(mid-l+1);
            tree[rson(x)].mark+=tree[x].mark;
            tree[rson(x)].v+=tree[x].mark*(r-mid);
            tree[x].mark=0;
        }
    }
    void update(int &x,int val,int L,int R,int l,int r){
        if(!x) x=++ptr;
        if(L<=l&&R>=r){
            tree[x].mark+=val;
            tree[x].v+=val*(r-l+1);
            return;
        }
        push_down(x,l,r);
        int mid=(l+r)>>1;
        if(L<=mid) update(tree[x].ls,val,L,R,l,mid);
        if(R>mid) update(tree[x].rs,val,L,R,mid+1,r);
        push_up(x);
    } 
    ll query(int x,int L,int R,int l,int r){
        if(!x) return 0;
        if(L<=l&&R>=r){
            return tree[x].v;
        } 
        push_down(x,l,r); 
        int mid=(l+r)>>1;
        ll ans=0;
        if(L<=mid) ans+=query(tree[x].ls,L,R,l,mid);
        if(R>mid) ans+=query(tree[x].rs,L,R,mid+1,r);
        return ans;
    } 
#undef lson
#undef rson
}T1;

struct segment_tree_val{//外层权值线段树 
    struct node{
        int l;
        int r;
        int root;
    }tree[maxn*4+5];
    void build(int x,int l,int r){
        tree[x].l=l;
        tree[x].r=r;
        if(l==r) return;
        int mid=(l+r)>>1;
        build(x<<1,l,mid);
        build(x<<1|1,mid+1,r);
    }
    void update(int x,int val,int L,int R){ //找到包含val的权值区间，在权值区间对应的内层线段树上区间修改 
        T1.update(tree[x].root,1,L,R,1,n);
        if(tree[x].l==tree[x].r) return;
        int mid=(tree[x].l+tree[x].r)>>1;
        if(val<=mid) update(x<<1,val,L,R);
        else update(x<<1|1,val,L,R); 
    }
    int query(int x,int k,int L,int R){
        if(tree[x].l==tree[x].r) return tree[x].l; 
        //注意是查询第k大 
        ll cnt=T1.query(tree[x<<1|1].root,L,R,1,n);//查询右子结点权值区间[l,mid]中,位置落在[L,R]内的有多少个 
        if(k<=cnt) return query(x<<1|1,k,L,R);
        else return query(x<<1,k-cnt,L,R); 
    }
}T2;

struct oper{
    int type;
    int a;
    int b;
    int c;
}q[maxm+5];
int dn;
int tmp[maxm+5];
int main(){
    qread(n);
    qread(m);
    for(int i=1;i<=m;i++){
        qread(q[i].type);
        qread(q[i].a);
        qread(q[i].b);
        qread(q[i].c);
        if(q[i].type==1) tmp[++dn]=q[i].c;
    }
    sort(tmp+1,tmp+1+dn);
    dn=unique(tmp+1,tmp+1+dn)-tmp-1;
    for(int i=1;i<=m;i++){
        if(q[i].type==1) q[i].c=lower_bound(tmp+1,tmp+1+dn,q[i].c)-tmp;
    } 
    T2.build(1,1,dn);
    for(int i=1;i<=m;i++){
        if(q[i].type==1){
            T2.update(1,q[i].c,q[i].a,q[i].b);
        }else{
            qprint(tmp[T2.query(1,q[i].c,q[i].a,q[i].b)]);//由于离散化过，tmp[x]才是真实的值 
            putchar('\n'); 
        }
    }
}