Gym102586B Evacuation

Evacuation

There are $N+2$ towns in a straight line. The towns are numbered from $0$ through $N+1$ from left to right. Each town $i$ ($1 \leq i \leq N$) has a shelter which can contain up to $A_i$ people.

Currently, $S$ people are traveling together and visiting one of the towns. However, you don't know which town they are visiting.

You just got to know that there are $Q$ meteorites that can hit the towns. The $i$-th meteorite may strike towns $L_i,L_i+1,\cdots,R_i$. To ensure the safety of the travelers, for each meteorite, you want to calculate the evacuation cost.

The evacuation cost for a meteorite is calculated as follows:

• The evacuation cost is the minimum total cost needed to make all travelers safe no matter which town they are visiting.

• A person is safe if he/she is in a shelter or a town outside the effect of the meteorite.

• It takes $1$ unit cost to move one person to an adjacent town (two towns are adjacent iff their numbers differ by exactly $1$).
  Note that only moving people costs money, and other actions (like accommodating people in a shelter) don't. It is guaranteed that towns $0$ and $N+1$ will never be affected by meteorites, so it is always possible to make all travelers safe.

Write a program that, for each meteorite, calculates the evacuation cost for that meteorite.

$1 \leq N,Q \leq 2 \times 10^5$

题解

有一个长度为$n$的序列$a$和一个常数$s$.

$q$次询问, 每次给一个区间$[l,r]$, 询问$\max_{l≤x≤r}f(l,r,x)$的值.

其中$f(l,r, x)$被定义为, 对于所有满足以下条件的序列$b$,

• $\sum^n_{i=1} b_i = s$

• $∀l ≤ i ≤ r, b_i ≤ a_i$

$\sum^n_{i=1} |i − x|b_i$的最小可能值

https://www.cnblogs.com/cjoierShiina-Mashiro/p/12800124.html

对于询问$f(l,r,x)$，如果我们固定了$x$，那么构造最优的$b$的方法是很简单的：

从小到大枚举$i\in[0,+\infty)$，尽可能地让$b_{x-i},b_{x+i}$大。

如果在$S$还没有放完的时候遇到了一个$pos\notin[l,r]$，那么把剩下的$S$全部放到$b_{pos}$就行了。

若$x\in[l,mid]$那么$pos=l-1$，若$x\in(mid,r]$那么$pos=r+1$。

不难发现$x\in[l,mid]$时$f(l,r,x)$与$r$无关，$x\in(mid,r]$时$f(l,r,x)$与$l$无关。

因此设$f(l,r,x)=\begin{cases}f(l,x)&x\in[l,mid]\\g(r,x)&x\in(mid,r]\end{cases}$

预处理$\{a_i\}$的前缀和以及$\{ia_i\}$的前缀和之后可以$O(1)$的求出单点$f,g$。

观察可得$f,g$满足四边形不等式。

考虑离线所有询问，对于一个形如$\max\limits_{x\in[L,R]}(f\text{ or }g)(r,x)$的询问，将其拆分成线段树上的$O(\log n)$个区间，最后对线段树的每个节点
用决策单调性优化暴力枚举$x$计算即可。

时间复杂度是$O(n\log^2n)$，利用基数排序可以做到$O(n\log n)$。

CO int N=2e5+10;
int n,q;
int64 s,s1[N],s2[N],ans[N];
int bound[N];
 
IN int64 sum(int64*s,int l,int r){
	return s[r]-s[l-1];
}
int find(int p){
	int l=0,r=min(n-p,p-1);
	while(l<r){
		int mid=(l+r+1)>>1;
		sum(s1,p-mid,p+mid)<=s?l=mid:r=mid-1;
	}
	return l;
}
 
IN int64 calc(int p,int len){
	if(len<=0) return 0;
	return sum(s1,p-len,p-1)*p-sum(s2,p-len,p-1)+sum(s2,p+1,p+len)-sum(s1,p+1,p+len)*p;
}
IN int64 calc(int o,int p,int lim){
	int f=!o?-1:1;
	int t=!o?max(lim-1,p-bound[p]-1):min(lim+1,p+bound[p]+1);
	return calc(p,f*(t-p)-1)+(s-(!o?sum(s1,t+1,2*p-(t+1)):sum(s1,2*p-(t-1),t-1)))*f*(t-p);
}
 
void solve(int o,int l,int r,vector<pair<int,int> >&q){
	if(q.empty()) return;
	if(l==r){
		for(CO pair<int,int>&a:q)
			ans[a.second]=max(ans[a.second],calc(o,l,a.first));
		return;
	}
	vector<pair<int,int> > ql,qr;
	int mid=q.size()/2,pos=0;
	int64 tmp,mx=0;
	for(int i=0;i<(int)q.size();++i)if(i!=mid)
		i<mid?ql.push_back(q[i]):qr.push_back(q[i]);
	for(int i=l;i<=r;++i)
		if((tmp=calc(o,i,q[mid].first))>mx) mx=tmp,pos=i;
	ans[q[mid].second]=max(ans[q[mid].second],mx);
	solve(o,l,pos,ql),solve(o,pos,r,qr);
}
 
vector<pair<int,int> > trans[4*N][2];
 
#define lc (x<<1)
#define rc (x<<1|1)
#define mid ((l+r)>>1)
void insert(int x,int l,int r,int ql,int qr,int o,int id){
	if(ql>qr) return;
	if(ql<=l and r<=qr) return trans[x][o].push_back({!o?ql:qr,id});
	if(ql<=mid) insert(lc,l,mid,ql,qr,o,id);
	if(qr>mid) insert(rc,mid+1,r,ql,qr,o,id);
}
void query(int x,int l,int r){
	for(int i=0;i<2;++i){
		sort(trans[x][i].begin(),trans[x][i].end());
		solve(i,l,r,trans[x][i]);
	}
	if(l==r) return;
	query(lc,l,mid),query(rc,mid+1,r);
}
#undef lc
#undef rc
#undef mid
 
int main(){
	read(n),read(s);
	for(int i=1;i<=n;++i){
		int64 x=read<int64>();
		s1[i]=s1[i-1]+x,s2[i]=s2[i-1]+x*i;
	}
	for(int i=1;i<=n;++i) bound[i]=find(i);
	read(q);
	for(int i=1;i<=q;++i){
		int l=read<int>(),r=read<int>(),mid=(l+r)>>1;
		insert(1,1,n,l,mid,0,i),insert(1,1,n,mid+1,r,1,i);
	}
	query(1,1,n);
	for(int i=1;i<=q;++i) write(ans[i],'\n');
	return 0;
}