CF1539F Strange Array

这玩意为啥是紫。

考虑对每个小于(x)的数值设为1，大于(x)的数值设为-1.
那么对于答案要求的就是绝对值最大的连续段。
线段树是很显然的。

考虑我们不能对每个数都进行一遍重构，这样就退化到了(O(n^2log))
我们对每个数的权值排序，那么更改操作变成了(O(nlog))

然后我们用线段树维护前缀和就好了。

CF1539F Strange Array

#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
#define N 200005

ll n;
ll num[N];

struct P{int v,to;}e[N];

bool operator < (P a,P b){
	return a.v < b.v;
}

struct Q{ll mx,mn,tag,s;Q(){mx = 0;mn = 0;tag = 0;}}t[N << 1];//维护前缀和，支持区间加，区间min，区间max 

#define l(x) (x << 1)
#define r(x) (x << 1 | 1)
#define mid ((l + r) >> 1)

inline void pushdown(int u){
	t[l(u)].mn += t[u].tag;
	t[r(u)].mn += t[u].tag;
	t[l(u)].mx += t[u].tag;
	t[r(u)].mx += t[u].tag;
	t[l(u)].tag += t[u].tag;
	t[r(u)].tag += t[u].tag;
	t[u].tag = 0;
} 

inline void up(int u){
	t[u].mx = std::max(t[l(u)].mx,t[r(u)].mx);
	t[u].mn = std::min(t[l(u)].mn,t[r(u)].mn);
}

ll l = 1;

inline void change(int u,int l,int r,int tl,int tr,int p){
	pushdown(u);
	if(tl <= l && r <= tr){
		t[u].mx += p;
		t[u].mn += p;
		t[u].tag += p;
//		std::cout<<u<<"mx:"<<t[u].mx<<"mi:"<<t[u].mn<<" "<<l<<" "<<r<<" "<<tl<<" "<<tr<<" "<<p<<std::endl; 
		return;
	}
	if(tl <= mid)
	change(l(u),l,mid,tl,tr,p);
	if(tr > mid)
	change(r(u),mid + 1,r,tl,tr,p);
	up(u);
//	std::cout<<u<<"mx:"<<t[u].mx<<"mi:"<<t[u].mn<<" "<<l<<" "<<r<<" "<<tl<<" "<<tr<<" "<<p<<std::endl; 	
} 

inline ll qx(int u,int l,int r,int tl,int tr){
	if(tr == 0)
	return 0;
	pushdown(u);
	ll ans = -0x3f3f3f3f;
	if(tl <= l && r <= tr)
	return t[u].mx;
	if(tl <= mid)
	ans = std::max(ans,qx(l(u),l,mid,tl,tr));
	if(tr > mid)
	ans = std::max(ans,qx(r(u),mid + 1,r,tl,tr));
	return ans; 
}

inline ll qi(int u,int l,int r,int tl,int tr){
//	std::cout<<u<<" "<<t[u].mn<<" "<<l<<" "<<r<<" "<<tl<<" "<<tr<<std::endl; 
	if(tr == 0)
	return 0;
	pushdown(u);
	ll ans = 0x3f3f3f3f;
	if(tl <= l && r <= tr)
	return t[u].mn;
	if(tl <= mid)
	ans = std::min(ans,qi(l(u),l,mid,tl,tr));
	if(tr > mid)
	ans = std::min(ans,qi(r(u),mid + 1,r,tl,tr));
	return ans; 
}

ll f[N],fa[N];

int main(){
	scanf("%lld",&n);
	for(int i = 1;i <= n;++i)
	scanf("%lld",&num[i]);
	for(int i = 1;i <= n;++i)
	e[i].v = num[i],e[i].to = i;
	for(int i = 1;i <= n;++i)
	change(1,1,n,i,n,1);
	std::sort(e + 1,e + n + 1);
	for(int i = 1;i <= n;++i){
		ll now = e[i].to;
		std::cout<<e[i].v<<" "<<e[i].to<<":"<<std::endl;
		change(1,1,n,e[i].to,n,-1);
		while(e[l].v <= e[i].v && l < i){
//			std::cout<<l<<" "<<std::endl;
			change(1,1,n,e[l].to,n,-2);
			l ++ ;
		}
		l = i + 1;
		while(e[l].v == e[i].v){
			change(1,1,n,e[l].to,n,-2)
			++l;
			std::cout<<l<<std::endl;
		}
		l -- ;
//		std::cout<<qx(1,1,n,now,n)<<" "<<std::min(qi(1,1,n,1,now - 1),(ll)0)<<std::endl;
//		std::cout<<qi(1,1,n,now,n)<<" "<<std::max(qx(1,1,n,1,now - 1),(ll)0)<<std::endl;
		ll ans = std::max((std::abs(qx(1,1,n,now,n) - std::min(qi(1,1,n,1,now - 1),(ll)0)) + 1),std::abs(qi(1,1,n,now,n) - std::max(qx(1,1,n,1,now - 1),(ll)0)));
		f[now] = ans / 2;
		change(1,1,n,e[i].to,n,1);
		while(l > i){
			change(1,1,n,e[l].to,n,2);
			
		} 		
	}
	for(int i = 1;i <= n;++i)
	std::cout<<f[i]<<" ";
}