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. Ruby 基础教程1-3

    1.命令行参数ARGV[] 2.文件读取 file=File.open(filename)    text=file.read  print text file.close 一次读取所有内容耗内存,耗 ...

  2. hdu1061Rightmost Digit(快速幂取余)

    Rightmost Digit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

  3. 怎样安装JMeter

    JMeter有图形界面, 而且支持中文! JMeter官网地址: http://jmeter.apache.org/ 点击左上角的下载: 点击下面的.zip后缀的压缩包: 解压到本地: JMeter目 ...

  4. 感觉总结了一切python常见知识点,可直接运行版本

    #encoding=utf-8#http://python.jobbole.com/85231/#作用域a=1def A(a): a=2 print 'A:',a def B(): print 'B: ...

  5. lintcode First Unique Number In Stream

    First Unique Number In Stream 描述: Given a continuous stream of numbers, write a function that return ...

  6. 【转】: 探索Lua5.2内部实现:虚拟机指令(3) Upvalues & Globals

    在编译期,如果要访问变量a时,会依照以下的顺序决定变量a的类型: a是当前函数的local变量 a是外层函数的local变量,那么a是当前函数的upvalue a是全局变量 local变量本身就存在于 ...

  7. leetcode个人题解——#5 Container with most water

    class Solution { public: string longestPalindrome(string s) { int length = s.length(); ) return s; ; ...

  8. ServiceStack.Ormlit 使用Insert的时候自增列不会被赋值

    Insert签名是这样的,将第2个参数设置为true就会返回刚插入的自增列ID了,然后可以手工赋值到对象上面去 public static long Insert<T>(this IDbC ...

  9. 【转载】inotify+rsync实时同步 解决同步慢问题 (转载备记)

    原文地址:http://www.ttlsa.com/web/let-infotify-rsync-fast/ 背景 我们公司在用inotify+rsync做实时同步,来解决分布式集群文件一致性的问题. ...

  10. P4编程环境搭建遇到的问题与解决方法

    在经历了无数的折腾之后,算是折腾,最后采用的是陈翔学长的脚本加上可爱的shell调整装好的. 链接:p4Install 也许是ubuntu18.04的问题,也有可能是我自己把这个系统折腾的有点杂乱的原 ...