Necklace

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2083    Accepted Submission(s): 747

Problem Description
Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value look the same, so if two or more balls have the same beautiful value, we just count it once. We define the beautiful value of some interval [x,y] as F(x,y). F(x,y) is calculated as the sum of the beautiful value from the xth ball to the yth ball and the same value is ONLY COUNTED ONCE. For example, if the necklace is 1 1 1 2 3 1, we have F(1,3)=1, F(2,4)=3, F(2,6)=6.

Now Mery thinks the necklace is too long. She plans to take some continuous part of the necklace to build a new one. She wants to know each of the beautiful value of M continuous parts of the necklace. She will give you M intervals [L,R] (1<=L<=R<=N) and you must tell her F(L,R) of them.

 
Input
The first line is T(T<=10), representing the number of test cases.
  For each case, the first line is a number N,1 <=N <=50000, indicating the number of the magic balls. The second line contains N non-negative integer numbers not greater 1000000, representing the beautiful value of the N balls. The third line has a number M, 1 <=M <=200000, meaning the nunber of the queries. Each of the next M lines contains L and R, the query.
 
Output
For each query, output a line contains an integer number, representing the result of the query.
 
Sample Input
2
6
1 2 3 4 3 5
3
1 2
3 5
2 6
6
1 1 1 2 3 5
3
1 1
2 4
3 5
 
Sample Output
3
7
14
1
3
6
 
Source
 
Recommend
lcy
 
  1. /*
  2. 题意为查找区间去重后的和
  3. 用树状数组离线处理
  4. 将所有查询以右端点从小到大排序
  5. 按此顺序边去重边查询
  6. 前面的去重就不会影响到后面的结果了
  7. */
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm> using namespace std; const int N=;
const int M=; struct node{
int l,r;
int id;
}q[M]; int n,m,val[N],pre[N],loc[];
long long arr[N],res[M]; int lowbit(int x){
return x&(-x);
} void update(int i,int x){
while(i<=n){
arr[i]+=x;
i+=lowbit(i);
}
} long long Sum(int i){
long long ans=;
while(i>){
ans+=arr[i];
i-=lowbit(i);
}
return ans;
} bool cmp(node a,node b){
return a.r<b.r;
} int main(){ //freopen("input.txt","r",stdin); int t;
scanf("%d",&t);
while(t--){
memset(arr,,sizeof(arr));
memset(loc,-,sizeof(loc));
scanf("%d",&n);
for(int i=;i<=n;i++){
scanf("%d",&val[i]);
pre[i]=loc[val[i]];
loc[val[i]]=i;
update(i,val[i]);
}
scanf("%d",&m);
for(int i=;i<=m;i++){
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
sort(q+,q++m,cmp);
int r=;
for(int i=;i<=m;i++){
for(int j=r+;j<=q[i].r;j++)
if(pre[j]!=-)
update(pre[j],-val[j]);
r=q[i].r;
res[q[i].id]=Sum(q[i].r)-Sum(q[i].l-);
}
for(int i=;i<=m;i++)
printf("%I64d\n",res[i]);
}
return ;
}

线段树:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map> using namespace std; const int N=; //#define L(rt) (rt<<1)
//#define R(rt) (rt<<1|1) #define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1 struct Tree{
int l,r;
int id;
}q[N<<]; map<int,int> mp;
int n,m,a[N];
long long sum[N<<],res[N<<]; void PushUp(int rt){
sum[rt]=sum[rt<<]+sum[rt<<|];
} void update(int id,int val,int l,int r,int rt){
if(l==r){
sum[rt]+=val;
return ;
}
int mid=(l+r)>>;
if(id<=mid)
update(id,val,lson);
else
update(id,val,rson);
PushUp(rt);
} long long query(int L,int R,int l,int r,int rt){
if(L<=l && R>=r)
return sum[rt];
int mid=(l+r)>>;
long long ans=;
if(L<=mid)
ans+=query(L,R,lson);
if(R>mid)
ans+=query(L,R,rson);
return ans;
} int cmp(Tree a,Tree b){
return a.r<b.r;
} int main(){ //freopen("input.txt","r",stdin); int t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
scanf("%d",&m);
for(int i=;i<=m;i++){
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
sort(q+,q++m,cmp);
mp.clear();
memset(sum,,sizeof(sum));
int r=;
for(int i=;i<=m;i++){
for(int j=r+;j<=q[i].r;j++){
if(mp[a[j]])
update(mp[a[j]],-a[j],,n,);
update(j,a[j],,n,);
mp[a[j]]=j;
r=q[i].r;
}
res[q[i].id]=query(q[i].l,q[i].r,,n,);
}
for(int i=;i<=m;i++)
printf("%I64d\n",res[i]);
}
return ;
}

HDU 3874 Necklace (树状数组 | 线段树 的离线处理)的更多相关文章

  1. 树状数组 && 线段树应用 -- 求逆序数

    参考:算法学习(二)——树状数组求逆序数 .线段树或树状数组求逆序数(附例题) 应用树状数组 || 线段树求逆序数是一种很巧妙的技巧,这个技巧的关键在于如何把原来单纯的求区间和操作转换为 求小于等于a ...

  2. hdu1394(枚举/树状数组/线段树单点更新&区间求和)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1394 题意:给出一个循环数组,求其逆序对最少为多少: 思路:对于逆序对: 交换两个相邻数,逆序数 +1 ...

  3. 洛谷P2414 阿狸的打字机 [NOI2011] AC自动机+树状数组/线段树

    正解:AC自动机+树状数组/线段树 解题报告: 传送门! 这道题,首先想到暴力思路还是不难的,首先看到y有那么多个,菜鸡如我还不怎么会可持久化之类的,那就直接排个序什么的然后按顺序做就好,这样听说有7 ...

  4. hdu 1166:敌兵布阵(树状数组 / 线段树,入门练习题)

    敌兵布阵 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submis ...

  5. hdu 5147 Sequence II【树状数组/线段树】

    Sequence IITime Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem ...

  6. hdu 3966 Aragorn's Story(树链剖分+树状数组/线段树)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3966 题意: 给出一棵树,并给定各个点权的值,然后有3种操作: I C1 C2 K: 把C1与C2的路 ...

  7. hdu 1166 敌兵布阵——(区间和)树状数组/线段树

    pid=1166">here:http://acm.hdu.edu.cn/showproblem.php?pid=1166 Input 第一行一个整数T.表示有T组数据. 每组数据第一 ...

  8. HDU 1166 敌兵布阵 树状数组||线段树

    http://acm.hdu.edu.cn/showproblem.php?pid=1166 题目大意: 给定n个数的区间N<=50000,还有Q个询问(Q<=40000)求区间和. 每个 ...

  9. HDU 3303 Harmony Forever 前缀和+树状数组||线段树

    Problem Description We believe that every inhabitant of this universe eventually will find a way to ...

  10. hdu 1166 树状数组(线段树)

    敌兵布阵 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submis ...

随机推荐

  1. IT人士感言2(转)

    01. 自己的户口档案.养老保险.医疗保险.住房公积金一定要保管好.由于程序员行业每年跳槽一次,我不隐瞒大家,我至少换过5个以上的单位,这期间跳来跳去,甚至是城市都换过3个.还好户口没丢掉,其他都已经 ...

  2. html写法对gzip压缩率的影响

    前几天在群里看到小杜分享一篇文章,<html写法对gzip压缩率的影响>,为此我也对这点分析了一下.不知道大家有没有看过这文章,作者是来自微博懒懒交流会,其内容我这里先简述一下. Gzip ...

  3. DOS命令大全(经典收藏)

    net use \\ip\ipc$ " " /user:" " 建立IPC空链接 net use \\ip\ipc$ "密码" /user: ...

  4. Remote Desktop Connection from Windows 7 to Ubuntu 12.04

    $sudo apt-get install xrdp $cd ~ $sudo vim .xsession gnome-session --session=ubuntu-2d 在windows下进行远程 ...

  5. 安装xubuntu时遇到的一些问题

    1  下载地址 http://www.linuxdown.net/ 2  选择虚拟机 VirtualBox 3  安装步骤 http://www.cnblogs.com/zhcncn/p/398730 ...

  6. SOA问题处理

    R12.1: How To Generate SOA Log For Debugging SOA Provider Issues (文档 ID 828753.1) 转到底部 In this Docum ...

  7. HTTP协议状态码详解(HTTP Status Code)

    转自:http://www.cnblogs.com/shanyou/archive/2012/05/06/2486134.html 使用ASP.NET/PHP/JSP 或者javascript都会用到 ...

  8. SQLLDR记录数与文本记录数比较

    我们平时都用sqlldr进行将文本数据加载到数据库,但是有时候由于数据问题导致入库率不能达到100%,因此我们要检测是否存在不能入库的数据记录.以下shell脚本就是统计文本中记录数和数据库中记录数是 ...

  9. ASP.NET MVC学习之模型模板篇

    一.前言 如果你使用ASP.NET MVC制作后台一定会爱上它的EditorForModal.DisplayForModal和LabelForModal方法,因为这些方法可以将模型直接变成对应的标签, ...

  10. ubuntu apache2 wsgi 部署django

    入题 分为如下几步 1.安装python 2.安装django 3.安装wsgi,如有问题请参照上一篇 ubuntu 编译安装 mod_wsgi 4.与apache集成这里主要讲这部分 环境apach ...