Recursive Queries CodeForces - 1117G (线段树)

题面:

刚开始想复杂了, 还以为是个笛卡尔树....

实际上我们发现, 对于询问(l,r)每个点的贡献是$min(r,R[i])-max(l,L[i])+1$

数据范围比较大在线树套树的话明显过不了, 还是想离线的算法好了, 只考虑求$\sum min(r,R[i])$, 对于$\sum max(l,L[i])$同理

将询问按$l$从大到小排, 将点$x$的贡献转化为$[x,R[x]-1]$区间加等差, $[R[x],n]$区间加$R[x]$, 这样$\sum min(r,R[i])$就变成对位置$r$单点求和了

离线的代码如下

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head

const int N = 1e6+10;
int n, q;
int a[N], L[N], R[N];
struct _ {int l,r,id;} e[N];
int v1[N<<2], t1[N<<2];
ll v2[N<<2], t2[N<<2], ans[N];

void pd1(int o) {
	if (t1[o]) {
		v1[lc]+=t1[o];
		v1[rc]+=t1[o];
		t1[lc]+=t1[o];
		t1[rc]+=t1[o];
		t1[o]=0;
	}
}
void pd2(int o) {
	if (t2[o]) {
		v2[lc]+=t2[o];
		v2[rc]+=t2[o];
		t2[lc]+=t2[o];
		t2[rc]+=t2[o];
		t2[o]=0;
	}
}
void upd1(int o, int l, int r, int ql, int qr, int v) {
	if (ql<=l&&r<=qr) return v1[o]+=v,t1[o]+=v,void();
	pd1(o);
	if (mid>=ql) upd1(ls,ql,qr,v);
	if (mid<qr) upd1(rs,ql,qr,v);
}
void upd2(int o, int l, int r, int ql, int qr, int v) {
	if (ql<=l&&r<=qr) return v2[o]+=v,t2[o]+=v,void();
	pd2(o);
	if (mid>=ql) upd2(ls,ql,qr,v);
	if (mid<qr) upd2(rs,ql,qr,v);
}
ll qry1(int o, int l, int r, int x) {
	if (l==r) return (ll)l*v1[o];
	pd1(o);
	if (mid>=x) return qry1(ls,x);
	return qry1(rs,x);
}
ll qry2(int o, int l, int r, int x) {
	if (l==r) return (ll)v2[o];
	pd2(o);
	if (mid>=x) return qry2(ls,x);
	return qry2(rs,x);
}

int main() {
	scanf("%d%d", &n, &q);
	REP(i,1,n) scanf("%d", a+i);
	REP(i,1,n) {
		L[i] = i-1;
		while (L[i]&&a[i]>a[L[i]]) L[i]=L[L[i]];
	}
	PER(i,1,n) {
		R[i] = i+1;
		while (R[i]<=n&&a[i]>a[R[i]]) R[i]=R[R[i]];
	}
	REP(i,1,n) ++L[i],--R[i];
	REP(i,1,q) scanf("%d", &e[i].l);
	REP(i,1,q) scanf("%d", &e[i].r),e[i].id=i;
	sort(e+1,e+1+q,[](_ a,_ b){return a.l>b.l;});
	int now = n;
	REP(i,1,q) {
		while (now>=e[i].l) {
			if (now<=R[now]-1) upd1(1,1,n,now,R[now]-1,1);
			upd2(1,1,n,R[now],n,R[now]);
			--now;
		}
		ans[e[i].id] += qry1(1,1,n,e[i].r)+qry2(1,1,n,e[i].r)+e[i].r-e[i].l+1;
	}
	memset(v1,0,sizeof v1);
	memset(v2,0,sizeof v2);
	memset(t1,0,sizeof t1);
	memset(t2,0,sizeof t2);
	sort(e+1,e+1+q,[](_ a,_ b){return a.r<b.r;});
	now = 1;
	REP(i,1,q) {
		while (now<=e[i].r) {
			if (L[now]+1<=now) upd1(1,1,n,L[now]+1,now,1);
			upd2(1,1,n,1,L[now],L[now]);
			++now;
		}
		ans[e[i].id] -= qry1(1,1,n,e[i].l)+qry2(1,1,n,e[i].l);
	}
	REP(i,1,q) printf("%lld ", ans[i]);hr;
}

在线的代码如下, RE on test 8.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc TL[o]
#define rc TR[o]
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head

const int N = 1e6+10, M = 1e7+10;
int n, q, tot;
int a[N], L[N], R[N], ql[N], qr[N];
int cntL[N], cntR[N], sumL[N], sumR[N];
int TL[M], TR[M];
ll val[M];

void add(int &o, int l, int r, int x, int v) {
	if (!o) o=++tot;
	val[o] += v;
	if (l==r) return;
	if (mid>=x) add(ls,x,v);
	else add(rs,x,v);
}
ll query(int o, int l, int r, int ql, int qr) {
	if (!o) return 0;
	if (ql<=l&&r<=qr) return val[o];
	ll ans = 0;
	if (mid>=ql) ans += query(ls,ql,qr);
	if (mid<qr) ans += query(rs,ql,qr);
	return ans;
}

int main() {
	scanf("%d%d", &n, &q);
	REP(i,1,n) scanf("%d", a+i);
	REP(i,1,n) {
		L[i] = i-1;
		while (L[i]&&a[i]>a[L[i]]) L[i]=L[L[i]];
	}
	PER(i,1,n) {
		R[i] = i+1;
		while (R[i]<=n&&a[i]>a[R[i]]) R[i]=R[R[i]];
	}
	REP(i,1,n) ++L[i],--R[i];
	REP(i,1,n) {
		for (int j=R[i]; j<=n; j+=j&-j) {
			add(cntR[j],1,n,i,1);
			add(sumR[j],1,n,i,R[i]);
		}
		for (int j=L[i]; j; j^=j&-j) {
			add(cntL[j],1,n,i,1);
			add(sumL[j],1,n,i,L[i]);
		}
	}
	REP(i,1,q) scanf("%d", ql+i);
	REP(i,1,q) scanf("%d", qr+i);
	REP(i,1,q) {
		ll sumr=0,suml=0,cntr=0,cntl=0;
		int l=ql[i], r=qr[i];
		for (int j=r; j; j^=j&-j) {
			sumr += query(sumR[j],1,n,l,r);
			cntr += query(cntR[j],1,n,l,r);
		}
		for (int j=l; j<=n; j+=j&-j) {
			suml += query(sumL[j],1,n,l,r);
			cntl += query(cntL[j],1,n,l,r);
		}
		ll ans = (ll)(r-l+1-cntr)*r+sumr;
		ans -= (ll)(r-l+1-cntl)*l+suml;
		ans += r-l+1;
		printf("%lld ", ans);
	} hr;
}