CF1375H Set Merging

Description

You are given a permutation $ a_1, a_2, dots, a_n $ of numbers from $ 1 $ to $ n $ . Also, you have $ n $ sets $ S_1,S_2,dots, S_n $ , where $ S_i={a_i} $ . Lastly, you have a variable $ cnt $ , representing the current number of sets. Initially, $ cnt = n $ .

We define two kinds of functions on sets:

$ f(S)=minlimits_{uin S} u $ ;

$ g(S)=maxlimits_{uin S} u $ .

You can obtain a new set by merging two sets $ A $ and $ B $ , if they satisfy $ g(A) < f(B) $ (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

- $ cntgets cnt+1 $ ;
- $ S_{cnt}=S_ucup S_v $ , you are free to choose $ u $ and $ v $ for which $ 1le u, v < cnt $ and which satisfy $ g(S_u) < f(S_v) $ .

You are required to obtain some specific sets.

There are $ q $ requirements, each of which contains two integers $ l_i $ , $ r_i $ , which means that there must exist a set $ S_{k_i} $ ( $ k_i $ is the ID of the set, you should determine it) which equals $ {a_umid l_ileq uleq r_i} $ , which is, the set consisting of all $ a_i $ with indices between $ l_i $ and $ r_i $ .

In the end you must ensure that $ cntleq 2.2 imes 10^6 $ . Note that you don&#039;t have to minimize $ cnt $ . It is guaranteed that a solution under given constraints exists.

Solution

合并两个集合的要求是它们其中一个的最大值小于另一个的最小值，题中要求合并出一段区间

在权值线段树上每个节点维护该节点上的数的在原数组中的下标

每次合并出给定集合可以在权值线段树上爆搜，在每个节点找到在区间内的数合并即可

因为每个节点最多可以合并出的集合数为$len^2$，$len$为节点大小，可以用map维护每个节点已经合并出哪些区间的集合

在线段树较上层，限制时间复杂度的是每个节点的经过次数和，因为不可能每个节点的所有区间都会被合并出

在线段树较下层，限制时间复杂度的是每个节点的能合并出的集合数之和，因为当所有集合都被合并出就相当于$O(1)$回答了

解上层和下层的分界线在第几层

$$sum_{i=1}^x (2^{log n -i}frac{{2^i}^2}{2})=qspace 2^{log n-x}$$

解得$x=frac{log q}{2}$

将两层的时间复杂度上限算出，即将$x$回代

时间复杂度$O(nsqrt q)$

#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<utility>
#include<vector>
#include<cstdio>
#include<map>
using namespace std;
int n,pos[5005],ans[100005],tot,q;
map<pair<int,int>,int>mp[20005];
vector<int>ve[20005];
pair<int,int>S[2200005];
inline int read(){
    int f=1,w=0;
    char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')w=(w<<1)+(w<<3)+ch-'0',ch=getchar();
    return f*w;
}
void build(int i,int l,int r){
    for(int j=l;j<=r;j++)ve[i].push_back(pos[j]);
    sort(ve[i].begin(),ve[i].end());
    if(l==r)return;
    int mid=l+r>>1;
    build(i<<1,l,mid),build(i<<1|1,mid+1,r);
}
int query(int i,int l,int r,int L,int R){
    int left=lower_bound(ve[i].begin(),ve[i].end(),L)-ve[i].begin(),right=upper_bound(ve[i].begin(),ve[i].end(),R)-ve[i].begin()-1,mid=l+r>>1;
    if(right<left)return 0;
    if(right==left)return ve[i][left];
    if(mp[i].find(make_pair(left,right))!=mp[i].end())return mp[i][make_pair(left,right)];
    int lc=query(i<<1,l,mid,L,R),rc=query(i<<1|1,mid+1,r,L,R);
    if(!lc||!rc)return mp[i][make_pair(left,right)]=lc|rc;
    S[++tot]=make_pair(lc,rc);
    return mp[i][make_pair(left,right)]=tot;
}
int main(){
    tot=n=read(),q=read();
    for(int i=1;i<=n;i++)pos[read()]=i;
    build(1,1,n);
    for(int i=1;i<=q;i++){
        int l=read(),r=read();
        ans[i]=query(1,1,n,l,r);
    }
    printf("%d
",tot);
    for(int i=n+1;i<=tot;i++)printf("%d %d
",S[i].first,S[i].second);
    for(int i=1;i<=q;i++)printf("%d ",ans[i]);
    return 0;
}