笔记-Recursive Queries

Recursive Queries

---

\[m_{l,r}=\textrm{id}(\max_{i=l}^r a_i)\\
f(l,r)=
\begin{cases}
(r-l+1)+f(l,m_{l,r}-1)+f(m_{l,r}+1,r)&\textrm{if } l\le r\\
0&\textrm{else}\\
\end{cases}
\]

设 \(L_i=\max\{j\}(j<i,a_j>a_i)\)，\(R_i=\min\{j\}(j>i,a_j>a_i)\)。

考虑每个 \(i\in[l,r]\) 成为 \(m_{L_i+1,R_i-1}\) 时对答案的贡献：

\[\textrm{len}\Big([\max(l,L_i+1),\min(r,R_i-1)]\Big)
\]

\[\therefore f(l,r)=\sum_{i=l}^r\min(r,R_i-1)-\max(l,L_i+1)+1
\]

---

Code

#include <bits/stdc++.h>
using namespace std;

//Start
typedef long long ll;
typedef double db;
#define mp(a,b) make_pair(a,b)
#define x(a) a.first
#define y(a) a.second
#define b(a) a.begin()
#define e(a) a.end()
#define sz(a) int((a).size())
#define pb(a) push_back(a)
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;

//Data
const int N=1e6;
int n,m,a[N+7],ql[N+7],qr[N+7];
int l[N+7],r[N+7];
vector<pair<int,int> > st[N+7];
ll ans[N+7];
typedef vector<ll> bit;
bit cnt,sm;
void add(bit&c,int x,ll y){for(;x<sz(c);x+=x&-x) c[x]+=y;}
ll sum(bit&c,int x){ll y=0;for(;x;x-=x&-x) y+=c[x];return y;}
ll sum(bit&c,int x,int y){return sum(c,y)-sum(c,x-1);}

//Main
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<=m;i++) scanf("%d",&ql[i]),st[ql[i]-1].pb(mp(i,-1));
	for(int i=1;i<=m;i++) scanf("%d",&qr[i]),st[qr[i]].pb(mp(i,1));
	a[0]=a[n+1]=inf;
	for(int i=1;i<=n;i++){l[i]=i-1;while(a[l[i]]<a[i]) l[i]=l[l[i]];}
	for(int i=n;i>=1;i--){r[i]=i+1;while(a[r[i]]<a[i]) r[i]=r[r[i]];}
	for(int i=1;i<=n;i++) l[i]++,r[i]--;
	for(int i=1;i<=m;i++) ans[i]=qr[i]-ql[i]+1;
	cnt=sm=bit(n+7,0);
	for(int i=1;i<=n;i++){
		add(cnt,l[i],1),add(sm,l[i],l[i]);
		for(auto j:st[i]) ans[x(j)]-=(sum(sm,ql[x(j)],qr[x(j)])+sum(cnt,1,ql[x(j)]-1)*ql[x(j)])*y(j);
	}
	cnt=sm=bit(n+7,0);
	for(int i=1;i<=n;i++){
		add(cnt,r[i],1),add(sm,r[i],r[i]);
		for(auto j:st[i]) ans[x(j)]+=(sum(sm,ql[x(j)],qr[x(j)])+sum(cnt,qr[x(j)]+1,n)*qr[x(j)])*y(j);
	}
	for(int i=1;i<=m;i++) printf("%lld%c",ans[i],"\n "[i<m]);
	return 0;
}

---

祝大家学习愉快！