Alyona and towers CodeForces - 739C (线段树)

Posted 2021-11-26 uid001

大意: 给定序列, 要求实现区间加, 询问整个序列最长的先增后减的区间.

 

线段树维护左右两端递增,递减,先增后减的长度即可, 要注意严格递增, 合并时要注意相等的情况, 要注意相加会爆int.

 

#include <iostream>
#include <random>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#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‘
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;})
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;}
inline int rd() {int x=0;char p=getchar();while(p<‘0‘||p>‘9‘)p=getchar();while(p>=‘0‘&&p<=‘9‘)x=x*10+p-‘0‘,p=getchar();return x;}
//head



const int N = 3e5+10;
int n, m;
struct _ {
	ll l,r,tag;
	int len,La,Lb,Lab,Ra,Rb,Rab,ab;
	_ () {}
	_ (int x) {
		l=r=x,len=La=Lb=Lab=Ra=Rb=Rab=ab=1,tag=0;
	}
	void upd(ll x) {
		l+=x,r+=x,tag+=x;
	}
	_ operator + (const _ &rhs) const {
		_ ret;
		ret.l=l,ret.r=rhs.r;
		ret.len=len+rhs.len;
		ret.La = La+(La==len&&r<rhs.l?rhs.La:0);
		ret.Lb = Lb+(Lb==len&&r>rhs.l?rhs.Lb:0);
		ret.Lab = Lab;
		if (La==len) { 
			if (r<rhs.l) ret.Lab=La+rhs.Lab;
			else if (r>rhs.l) ret.Lab=La+rhs.Lb;
		}
		else if (Lab==len&&r>rhs.l) ret.Lab=Lab+rhs.Lb;
		ret.Ra = rhs.Ra+(rhs.Ra==rhs.len&&r<rhs.l?Ra:0);
		ret.Rb = rhs.Rb+(rhs.Rb==rhs.len&&r>rhs.l?Rb:0);
		ret.Rab = rhs.Rab;
		if (rhs.Rb==rhs.len) {
			if (r<rhs.l) ret.Rab=rhs.Rb+Ra;
			else if (r>rhs.l) ret.Rab=rhs.Rb+Rab;
		}
		else if (rhs.Rab==rhs.len&&r<rhs.l) ret.Rab=rhs.Rab+Ra;
		ret.ab = max(ab,rhs.ab);
		if (r!=rhs.l) ret.ab=max(ret.ab,Ra+rhs.Lb);
		if (r<rhs.l) ret.ab=max(ret.ab,Ra+rhs.Lab);
		if (r>rhs.l) ret.ab=max(ret.ab,Rab+rhs.Lb);
		ret.tag = 0;
		return ret;
	}
} tr[N<<2];
void build(int o, int l, int r) {
	if (l==r) tr[o]=_(rd());
	else build(ls),build(rs),tr[o]=tr[lc]+tr[rc];
}
void pd(int o) {
	if (tr[o].tag) {
		tr[lc].upd(tr[o].tag);
		tr[rc].upd(tr[o].tag);
		tr[o].tag=0;
	}
}
void update(int o, int l, int r, int ql, int qr, int v) {
	if (ql<=l&&r<=qr) return tr[o].upd(v);
	pd(o);
	if (mid>=ql) update(ls,ql,qr,v);
	if (mid<qr) update(rs,ql,qr,v);
	tr[o]=tr[lc]+tr[rc];
}
int main() {
	build(1,1,n=rd());
	m=rd();
	REP(i,1,m) {
		int l=rd(),r=rd(),d=rd();
		update(1,1,n,l,r,d);	
		printf("%d\n", tr[1].ab);
	}
}