区间连续不重复子段最大值,要维护历史的最大值和当前的最大值,打两个lazy,离线

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 150000
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define LL long long
using namespace std; typedef struct {
LL nmax,hmax,nlazy,hlazy;
}Tree;
Tree tree[maxn*]; typedef struct {
int l,r,id;
} Que;
Que que[maxn*]; LL num[maxn*],ans[maxn*];
int n,m,pre[maxn*]; int cmp(Que x,Que y)
{
if (x.r<y.r) return ;
return ;
} void update(int x)
{
tree[x].nmax=max(tree[x<<].nmax,tree[x<<|].nmax);
tree[x].hmax=max(tree[x<<].hmax,tree[x<<|].hmax);
} void pushdown(int x)
{
if (tree[x].hlazy) {
tree[x<<].hmax=max(tree[x<<].hmax,tree[x<<].nmax+tree[x].hlazy);
tree[x<<|].hmax=max(tree[x<<|].hmax,tree[x<<|].nmax+tree[x].hlazy);
tree[x<<].hlazy=max(tree[x<<].hlazy,tree[x].hlazy+tree[x<<].nlazy);
tree[x<<|].hlazy=max(tree[x<<|].hlazy,tree[x].hlazy+tree[x<<|].nlazy);
tree[x].hlazy=;
}
if (tree[x].nlazy) {
tree[x<<].nmax=tree[x<<].nmax+tree[x].nlazy;
tree[x<<|].nmax=tree[x<<|].nmax+tree[x].nlazy;
tree[x<<].nlazy+=tree[x].nlazy;
tree[x<<|].nlazy+=tree[x].nlazy;
tree[x].nlazy=;
}
} void change(int x,int l,int r,int ll,int rr,LL y)
{
if (ll<=l && r<=rr) {
tree[x].nlazy+=y;
tree[x].nmax+=y;
tree[x].hlazy=max(tree[x].hlazy,tree[x].nlazy);
tree[x].hmax=max(tree[x].nmax,tree[x].hmax);
return;
}
pushdown(x);
int mid=(l+r)>>;
if (ll<=mid) change(x<<,l,mid,ll,rr,y);
if (rr>mid) change(x<<|,mid+,r,ll,rr,y);
update(x);
} LL ask(int x,int l,int r,int ll,int rr)
{
// printf("%d %d %d %lld %lld\n",x,l,r,tree[x].hmax,tree[x].nmax);
if (ll<=l && r<=rr) return tree[x].hmax;
pushdown(x);
int mid=(l+r)>>;
if (rr<=mid) return ask(x<<,l,mid,ll,rr);
else
if (ll>mid) return ask(x<<|,mid+,r,ll,rr);
else
return max(ask(x<<,l,mid,ll,mid),ask(x<<|,mid+,r,mid+,rr));
} void build(int x,int l,int r)
{
tree[x].hlazy=tree[x].nlazy=;
if (l==r) {
scanf("%lld",&num[l]);
tree[x].hmax=tree[x].nmax=;
return;
}
int mid=(l+r)>>;
if (l<=mid) build(x<<,l,mid);
if (mid<r) build(x<<|,mid+,r);
update(x);
} int main()
{
scanf("%d",&n);
build(,,n);
scanf("%d",&m);
rep(i,,m-) {
scanf("%d %d",&que[i].l,&que[i].r);
que[i].id=i;
}
sort(que,que+m,cmp);
memset(pre,,sizeof(pre));
int now=;
rep(i,,n) {
change(,,n,pre[num[i]+maxn]+,i,num[i]);
pre[num[i]+maxn]=i;
while (now<=m && que[now].r==i) {
ans[que[now].id]=ask(,,n,que[now].l,que[now].r);
++now;
}
}
rep(i,,m-) printf("%lld\n",ans[i]);
return ;
}

【SPOJ - GSS2】Can you answer these queries II(线段树)的更多相关文章

  1. bzoj 2482: [Spoj GSS2] Can you answer these queries II 线段树

    2482: [Spoj1557] Can you answer these queries II Time Limit: 20 Sec  Memory Limit: 128 MBSubmit: 145 ...

  2. SPOJ GSS2 Can you answer these queries II ——线段树

    [题目分析] 线段树,好强! 首先从左往右依次扫描,线段树维护一下f[].f[i]表示从i到当前位置的和的值. 然后询问按照右端点排序,扫到一个位置,就相当于查询区间历史最值. 关于历史最值问题: 标 ...

  3. SPOJ 1557. Can you answer these queries II 线段树

    Can you answer these queries II Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 https://www.spoj.com/pr ...

  4. Spoj 1557 Can you answer these queries II 线段树 随意区间最大子段和 不反复数字

    题目链接:点击打开链接 每一个点都是最大值,把一整个序列和都压缩在一个点里. 1.普通的区间求和就是维护2个值,区间和Sum和延迟标志Lazy 2.Old 是该区间里出现过最大的Sum, Oldlaz ...

  5. SPOJ GSS2 - Can you answer these queries II(线段树 区间修改+区间查询)(后缀和)

    GSS2 - Can you answer these queries II #tree Being a completist and a simplist, kid Yang Zhe cannot ...

  6. spoj gss2 : Can you answer these queries II 离线&&线段树

    1557. Can you answer these queries II Problem code: GSS2 Being a completist and a simplist, kid Yang ...

  7. SPOJ GSS3 Can you answer these queries III[线段树]

    SPOJ - GSS3 Can you answer these queries III Description You are given a sequence A of N (N <= 50 ...

  8. 【BZOJ2482】[Spoj1557] Can you answer these queries II 线段树

    [BZOJ2482][Spoj1557] Can you answer these queries II Description 给定n个元素的序列. 给出m个询问:求l[i]~r[i]的最大子段和( ...

  9. SPOJ GSS2 Can you answer these queries II

    Time Limit: 1000MS   Memory Limit: 1572864KB   64bit IO Format: %lld & %llu Description Being a ...

  10. SPOJ GSS1 - Can you answer these queries I(线段树维护GSS)

    Can you answer these queries I SPOJ - GSS1 You are given a sequence A[1], A[2], -, A[N] . ( |A[i]| ≤ ...

随机推荐

  1. 单例模式之pymysql运用实例

    何为单例? 简单介绍一下下:单例是个什么鬼东西!!!! 单例模式含义] 单例模式是一种常用的软件设计模式.在它的核心结构中只包含一个被称为单例类的特殊类.通过单例模式可以保证系统中一个类只有一个实例而 ...

  2. ORB-SLAM(七)ORBextractor 特征提取

    该类中主要调用OpenCV中的函数,提取图像中特征点(关键点及其描述,描述子,以及图像金字塔) 参考TUM1.yaml文件中的参数,每一帧图像共提取1000个特征点,分布在金字塔8层中,层间尺度比例1 ...

  3. android 学习四 ContentProvider

    1.系统自带的许多数据(联系人,本地信息等)保存在sqllite数据库,然后封装成许多ContentProvider来供其他程序访问. 2.对sqllite数据库的操作,可以在命令行通过adb工具登录 ...

  4. Android 9 Pie震撼来袭 同步登陆WeTest

    WeTest 导读 2018年8月7日,Google对外发布最新 Android 9.0 正式版系统,并宣布系统版本Android P 被正式命名为代号“Pie”,最新系统已经正式推送包括谷歌Pixe ...

  5. 如何设置虚拟化的centos内、外网络通畅

    首先要去确定你的本机(本地物理机)是通过以太网(插网线)上网的,还是通过wifi上网的.这个很重要. 如果是通过以太网去上网,那么虚拟化出来的系统,网络配置应当选择桥接模式. 当然了,也不一定非要用桥 ...

  6. unity发布自定义分辨率

    如果你需要发布unity时想要使用自己设置的分辨率仅需要一下几个步骤: 打开Build Setting->PlayerSetting->Resolution and Presentatio ...

  7. [CodeForce721C]Journey

    题目描述 Recently Irina arrived to one of the most famous cities of Berland - the Berlatov city. There a ...

  8. 352[LeetCode] Data Stream as Disjoint Intervals

    Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen ...

  9. windows 7 安装docker

    下载docker-install.exe 下载链接 安装,一路next(安装完成后建议重启电脑) 点击桌面boot2docker Start图标,等待初始化,运行docker --version检验是 ...

  10. JQuery文本框验证

    <" CODEPAGE="936"%><!--#include file="conncon.asp"--><!--#in ...