CodeForces 173E Camping Groups 离线线段树 树状数组
Camping Groups
题目连接:
http://codeforces.com/problemset/problem/173/E
Description
A club wants to take its members camping. In order to organize the event better the club directors decided to partition the members into several groups.
Club member i has a responsibility value ri and an age value ai. A group is a non-empty subset of club members with one member known as group leader. A group leader should be one of the most responsible members of the group (his responsibility value is not less than responsibility of any other group member) and his age absolute difference with any other group member should not exceed k.
Some club members are friends and want to be in the same group. They also like their group to be as large as possible. Now you should write a program that answers a series of questions like "What's the largest size of a group containing club member x and club member y?". It's possible for x or y to be the group leader.
Input
The first line contains two integers n and k (2 ≤ n ≤ 105, 0 ≤ k ≤ 109) — the number of club members and the age restriction for one group.
The next line contains integer numbers r1, r2, ..., rn (1 ≤ ri ≤ 109) separated by space: ri denotes the i-th club member's responsibility. In the same way there are integers a1, a2, ..., an (1 ≤ ai ≤ 109) in the third line: ai denotes the i-th club member's age.
The next line contains an integer q denoting the number of questions that you should answer (1 ≤ q ≤ 105). The next q lines describe the questions. Each line contains two space-separated integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi) — the indices of the club members that should end up in the same group.
Output
For each question print the maximum size of the group in a line. If making such a group is impossible print -1 instead.
Sample Input
5 1
1 5 4 1 2
4 4 3 2 2
4
5 3
2 3
2 5
4 1
Sample Output
4
3
-1
4
Hint
题意
给你n个人,每个人都有两个属性点,ri和ai,表示这个人的领导值和年龄
然后可以让一个人当领导,他可以领导所有领导值小于等于他的人,年龄和他的之差不超过k的人
然后Q次询问
问你x,y所能够在的最大团队的大小是多少
题解:
离散化是显然的
首先,我们离线树状数组,按照领导值从小到大排序之后,年龄当成坐标,去维护每一个人当领导的时候能够领导多少个人
然后我们再把所有询问全部读入,把询问和每个人都扔到一个vector里面,按照领导值从大到小排序,这样显然先更新的点一定能够统领后面的人
然后再离线去维护询问就好了
代码
#include<bits/stdc++.h>
using namespace std;
struct node
{
int x,y,id;
int flag;
int ans;
}p[1300000],p2[1300000],p3[1300000];
int n,k;
vector<int> V;
map<int,int> H;
typedef int SgTreeDataType;
struct treenode
{
int L , R ;
SgTreeDataType sum , lazy;
};
struct Question
{
int l,r,lead,flag,ans;
};
vector<Question>T;
treenode tree[2300000];
inline void push_up(int o)
{
tree[o].sum = max(tree[2*o].sum , tree[2*o+1].sum);
}
inline void build_tree(int L , int R , int o)
{
tree[o].L = L , tree[o].R = R,tree[o].sum = tree[o].lazy = 0;
if (R > L)
{
int mid = (L+R) >> 1;
build_tree(L,mid,o*2);
build_tree(mid+1,R,o*2+1);
}
}
inline void updata(int QL,int QR,SgTreeDataType v,int o)
{
int L = tree[o].L , R = tree[o].R;
if (QL <= L && R <= QR) tree[o].sum = max(tree[o].sum,v);
else
{
int mid = (L+R)>>1;
if (QL <= mid) updata(QL,QR,v,o*2);
if (QR > mid) updata(QL,QR,v,o*2+1);
push_up(o);
}
}
inline SgTreeDataType query(int QL,int QR,int o)
{
int L = tree[o].L , R = tree[o].R;
if (QL <= L && R <= QR) return tree[o].sum;
else
{
int mid = (L+R)>>1;
SgTreeDataType res = 0;
if (QL <= mid) res = max(res,query(QL,QR,2*o));
if (QR > mid) res = max(res,query(QL,QR,2*o+1));
push_up(o);
return res;
}
}
struct Bit
{
vector<int> a;
int sz;
void init(int n)
{
sz=n;
for(int i=1;i<=n+5;i++)
a.push_back(0);
}
int lowbit(int x)
{
return x&(-x);
}
int query(int x)
{
if(x==0)return 0;
int ans = 0;
for(;x;x-=lowbit(x))ans+=a[x];
return ans;
}
void updata(int x,int v)
{
if(x==0)return;
for(;x<sz;x+=lowbit(x))
a[x]+=v;
}
}bit;
bool cmp(node a,node b)
{
if(a.x==b.x)return a.flag>b.flag;
return a.x<b.x;
}
bool cmp2(Question a,Question b)
{
if(a.lead == b.lead)
return a.flag<b.flag;
return a.lead>b.lead;
}
int ans[1300000];
int main()
{
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
scanf("%d",&p[i].x);
for(int i=1;i<=n;i++)
{
scanf("%d",&p[i].y);
V.push_back(p[i].y);
V.push_back(p[i].y-k);
V.push_back(p[i].y+k);
p[i].id=i;
p[i].flag=1;
}
sort(V.begin(),V.end());
V.erase(unique(V.begin(),V.end()),V.end());
for(int i=0;i<V.size();i++)
H[V[i]]=i+1;
for(int i=1;i<=n;i++)
p2[i]=p[i];
for(int i=1;i<=n;i++)
{
p2[i+n]=p[i];
p2[i+n].flag=-1;
}
sort(p2+1,p2+2*n+1,cmp);
bit.init(10*n);
for(int i=1;i<=2*n;i++)
{
if(p2[i].flag==1)
bit.updata(H[p2[i].y],1);
else
{
int sum = bit.query(H[p2[i].y+k])-bit.query(H[p2[i].y-k]-1);
p[p2[i].id].ans=sum;
}
}
for(int i=1;i<=n;i++)
{
Question now;
now.l = p[i].y;
now.r = p[i].y;
now.lead = p[i].x;
now.ans = p[i].ans;
now.flag = 0;
T.push_back(now);
}
int q;scanf("%d",&q);
for(int i=1;i<=q;i++)
{
int x,y;
scanf("%d%d",&x,&y);
if(p[x].y>p[y].y)swap(x,y);
Question now;
now.l = p[y].y-k;
now.r = p[x].y+k;
now.lead = max(p[x].x,p[y].x);
now.ans = 0;
now.flag = i;
T.push_back(now);
}
sort(T.begin(),T.end(),cmp2);
build_tree(1,5*n+5*q,1);
for(int i=0;i<T.size();i++)
{
if(T[i].flag==0)
updata(H[T[i].l],H[T[i].r],T[i].ans,1);
else
{
if(T[i].l>T[i].r)ans[T[i].flag]=-1;
else{
int ans1 = query(H[T[i].l],H[T[i].r],1);
ans[T[i].flag] = ans1;
}
}
}
for(int i=1;i<=q;i++)
{
if(ans[i]<2)
printf("-1\n");
else
printf("%d\n",ans[i]);
}
}
CodeForces 173E Camping Groups 离线线段树 树状数组的更多相关文章
- CodeForces - 1087F:Rock-Paper-Scissors Champion(set&数状数组)
n players are going to play a rock-paper-scissors tournament. As you probably know, in a one-on-one ...
- CodeForces -163E :e-Government (AC自动机+DFS序+树状数组)
The best programmers of Embezzland compete to develop a part of the project called "e-Governmen ...
- HDU 4638 Group (线段树 | 树状数组 + 离线处理)
Group Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submi ...
- Educational Codeforces Round 10 D. Nested Segments 离线树状数组 离散化
D. Nested Segments 题目连接: http://www.codeforces.com/contest/652/problem/D Description You are given n ...
- HDU 3874 Necklace (树状数组 | 线段树 的离线处理)
Necklace Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total S ...
- Codeforces Round #365 (Div. 2) D. Mishka and Interesting sum (离线树状数组+前缀xor)
题目链接:http://codeforces.com/contest/703/problem/D 给你n个数,m次查询,每次查询问你l到r之间出现偶数次的数字xor和是多少. 我们可以先预处理前缀和X ...
- Codeforces Gym 100114 H. Milestones 离线树状数组
H. Milestones Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100114 Descripti ...
- hdu 4288 离线线段树+间隔求和
Coder Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Su ...
- bzoj2333 离线 + 线段树
https://www.lydsy.com/JudgeOnline/problem.php?id=2333 有N个节点,标号从1到N,这N个节点一开始相互不连通.第i个节点的初始权值为a[i],接下来 ...
随机推荐
- jQuery遮罩插件jQuery.blockUI.js简介
利用Jquery.blockui.js创建可拖动.自定义内容的弹出层 利用Jquery.blockui.js创建可拖动.自定义内容的弹出层 目标 : 1 . 弹出层的内容可以自定义任意的HTML元素 ...
- HDU5828 Rikka with Sequence 线段树
分析:这个题和bc round 73应该是差不多的题,当时是zimpha巨出的,那个是取phi,这个是开根 吐槽:赛场上写的时候直接维护数值相同的区间,然后1A,结果赛后糖教一组数据给hack了,仰慕 ...
- Ioc注入方式写dubbo client(非set beans)
@Autowired注解的方式注解 Spring框架中进行注入式,使用@Autowired. @Autowired可以对成员变量.方法和构造函数进行标注,来完成自动装配的工作,这里必须明确:@Auto ...
- jQuery中的bind绑定事件与文本框改变事件的临时解决方法
暂时没有想到什么好的解决办法,我现在加了个浏览器判断非ie的话就注册blur事件,这样有个问题就是blur实在别的控件活动焦点的时候,txtStation控件注册的方法是为了填充它紧挨着的一个下拉列表 ...
- 用Asp.net实现简单的文字水印
用Asp.net实现简单的文字水印 经常看见MOP上有人贴那种动态的图片,就是把一个字符串作为参数传给一个动态网页,就会生成一个带有这个字符串的图片,这个叫做文字水印.像什么原来的熊猫系列,还有后来 ...
- 网络编程 --- URLConnection --- 读取服务器的数据 --- java
使用URLConnection类获取服务器的数据 抽象类URLConnection表示一个指向指定URL资源的活动连接,它是java协议处理器机制的一部分. URL对象的openConnection( ...
- 再来说说Activity
经过前面多天的了解,现在可以确信一点: activity提供了用户和程序交互的界面. 而且android里有四大组件:Activity,Service,BroadcastReceiver,Conten ...
- RESTful服务的版本管理经验 (转)
原文:RESTful服务的版本管理经验 最近,Howard Dierking将在设计NuGet API的下一个主要修订版(v3)时新学到的经验,与他在大约一年前的观念做了对比,并写道:使用服务器驱动的 ...
- log4j:ERROR LogMananger.repositorySelector was null likely due to error in class reloading, using NOPLoggerRepository.
The reason for the error is a new listener in Tomcat 6.0.24. You can fix this error by adding this l ...
- Type Encoding
[Type Encodings] The compiler encodes the return and argument types for each method in a character s ...