2015北京网络赛 J Clarke and puzzle 求五维偏序 分块+bitset
Clarke and puzzle
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://hihocoder.com/contest/acmicpc2015beijingonline/problem/10
Description
Kyle is a student of Programming Monkey Elementary School. Just as others, he is deeply concerned with his grades.
Last month, the school held an examination including five subjects, without any doubt, Kyle got a perfect score in every single subject.
There are n students took part in this examination(not including Kyle), and everyone got an integer between 1 to m as the score of one subject.
Now, looking at the grade table of these n students, Kyle wants to know how many students still did no better than him even if his scores are something else – Here, “no better” means there is no subject in which the student got strictly greater score than Kyle.
Input
There are multiple test cases.
The first line of the input contains an integer T (T <= 3) which means the number of test cases.
The first line of each test case contains two integers, n, m(n, m≤ 50,000), which are the number of students and the perfect score of each subject.
In the next n lines, each line consists of five integers, indicating a student’s scores.
Then one line follows. This line contains an integer q(q≤ 50,000) indicating the number of queries.
In the next q lines, each line contains five integers as well, representing a query. Each query indicates a set of scores, and for each query, you should figure out that if Kyle's grade is this set of scores, how many students still did no better than him. But for the sake of security, only the first query is in its original form, and other queries are encrypted. To decrypt a query, you must let each integer in the query do xor operation with the answer of last query. It's guaranteed that all the decrypted queries contain integers between 1 and 50000.
Output
For each test case, you should output q lines as the answer for all queries.
Sample Input
2
2 3
1 1 1 1 1
2 2 2 2 2
2
1 1 1 1 1
3 3 3 3 3
3 5
1 1 1 1 1
1 1 1 1 1
1 2 3 4 5
2
1 1 1 1 1
1 1 1 1 1
Sample Output
1
2
2
2
HINT
In case 1, there are two students with different scores and the scores of the first student (1, 1, 1, 1, 1) are not larger than the first query (1 1 1 1 1) in every subject, so the answer for this query is 1.
After having xor operation with the last answer 1, the second query (3,3,3,3,3) will be decrypted into (2, 2, 2, 2, 2). Because both students’ scores are no better than (2, 2, 2, 2, 2), so the answer for query 2 is 2.
题意
求五维偏序,强制在线
题解:
首先,如果维数低的话,就直接裸着跑kdtree就好了。
但是这个有五维,所以kdtree就不要想了
我写的是分块+bitset
我预处理5*sqrt(n)个bitset,bitset[i][j]表示在第i维,当前排名为j*sqrt(n)的时候的bitset的样子
那么每次询问的时候,我们就可以从最近的bitset里面转移过来就好了
对于每一个case,预处理的复杂度是O(n)的,询问的复杂度是5*sqrt(n),所以跑过去了~
代码:
//qscqesze
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1006
#define mod 1000000007
#define eps 1e-9
#define PI acos(-1)
const double EP = 1E- ;
int Num;
//const int inf=0x7fffffff;
const ll inf=;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************* bitset<> b[][];
bitset<> Ans[];
int now[];
struct node
{
int x,y;
};
bool cmp(node a,node b)
{
return a.x<b.x;
}
node a[][];
int l[],r[];
int belong[];
int main()
{
int t;scanf("%d",&t);
for(int cas=;cas<=t;cas++)
{
int n=read(),m=read();
for(int i=;i<=;i++)
for(int j=;j<;j++)
b[i][j].reset();
for(int i=;i<=n;i++)
{
for(int j=;j<=;j++)
{
a[j][i].x=read();
a[j][i].y=i;
}
}
for(int i=;i<=;i++)
sort(a[i]+,a[i]+n+,cmp);
int block = sqrt(n);
int num = n/block;
if(n%block)num++;
for(int i=;i<=num;i++)
l[i]=(i-)*block+,r[i]=i*block;
r[num]=n;
for(int i=;i<=n;i++)
belong[i]=(i-)/block+;
for(int i=;i<=;i++)
{
for(int j=;j<=num;j++)
{
b[i][j]|=b[i][j-];
for(int k=l[j];k<=r[j];k++)
b[i][j][a[i][k].y]=;
}
} int q=read();
int lastans = ;
while(q--)
{
for(int i=;i<=;i++)
now[i]=read();
for(int i=;i<=;i++)
now[i]^=lastans;
for(int i=;i<=;i++)
Ans[i].reset();
for(int i=;i<=;i++)
{
int L = ,R = n;
while(L<=R)
{
int mid = (L+R)/;
if(now[i]>=a[i][mid].x)
L = mid+;
else
R = mid-;
}
int p = L-;
if(p==)continue;
Ans[i]|=b[i][belong[p]-];
for(int j=l[belong[p]];j<=p;j++)
Ans[i][a[i][j].y]=;
}
Ans[] = Ans[]&Ans[]&Ans[]&Ans[]&Ans[];
lastans = Ans[].count();
printf("%d\n",lastans);
}
}
}
2015北京网络赛 J Clarke and puzzle 求五维偏序 分块+bitset的更多相关文章
- 2015北京网络赛 J Scores bitset+分块
2015北京网络赛 J Scores 题意:50000组5维数据,50000个询问,问有多少组每一维都不大于询问的数据 思路:赛时没有思路,后来看解题报告也因为智商太低看了半天看不懂.bitset之前 ...
- hihocoder 1236(2015北京网络赛 J题) 分块bitset乱搞题
题目大意: 每个人有五门课成绩,初始给定一部分学生的成绩,然后每次询问给出一个学生的成绩,希望知道在给定的一堆学生的成绩比这个学生每门都低或者相等的人数 因为强行要求在线查询,所以题目要求,每次当前给 ...
- 2015北京网络赛 Couple Trees 倍增算法
2015北京网络赛 Couple Trees 题意:两棵树,求不同树上两个节点的最近公共祖先 思路:比赛时看过的队伍不是很多,没有仔细想.今天补题才发现有个 倍增算法,自己竟然不知道. 解法来自 q ...
- 2015北京网络赛 D-The Celebration of Rabbits 动归+FWT
2015北京网络赛 D-The Celebration of Rabbits 题意: 给定四个正整数n, m, L, R (1≤n,m,L,R≤1000). 设a为一个长度为2n+1的序列. 设f(x ...
- hihocoder1236(北京网络赛J):scores 分块+bitset
北京网络赛的题- -.当时没思路,听大神们说是分块+bitset,想了一下发现确实可做,就试了一下,T了好多次终于过了 题意: 初始有n个人,每个人有五种能力值,现在有q个查询,每次查询给五个数代表查 ...
- Hiho coder 1236 2015 北京网络赛 Score
五维偏序..一开始被吓到了,后来知道了一种BITSET分块的方法,感觉非常不错. 呆马: #include <iostream> #include <cstdio> #incl ...
- acm 2015北京网络赛 F Couple Trees 树链剖分+主席树
Couple Trees Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://hihocoder.com/problemset/problem/123 ...
- 2015北京网络赛A题The Cats' Feeding Spots
题意:给你一百个点,找个以这些点为中心的最小的圆,使得这个圆恰好包含了n个点,而且这个圆的边界上并没有点 解题思路:暴力枚举每个点,求出每个点到其他点的距离,取第n大的点,判断一下. #include ...
- 2015北京网络赛 F Couple Trees 暴力倍增
Couple Trees Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://hihocoder.com/problemset/problem/123 ...
随机推荐
- Android开发之ListView-ArrayAdapter的使用
ArrayAdapter: ArrayAdapter<String>(Context context, int resource, int textViewResourceId, Stri ...
- Managed C++中使用Nullable<T>
非null值表示和C#的用法一样. Nullable<int> a = 1; null值的表示: Nullable<int> a = Nullable<int>() ...
- 十条常用nmap命令行格式
十条常用nmap命令行格式 ) 获取远程主机的系统类型及开放端口 nmap -sS -P0 -sV -O <target> 这里的 < target > 可以是单一 IP, 或 ...
- 数论/the second wave
扩展欧几里得算法. void exgcd(int a,int b,int&x,int&y){ if(!b) { x=1;y=0;return ; } exgcd(b,a%b,x,y); ...
- BZOJ2140: 稳定婚姻
题解: 题意就是求二分图的必须边. 我们有结论: 在残量网络上跑tarjan,对于一条边(u,v) 如果该边满流||scc[u]==scc[v],那么该边是可行边. 因为如果scc[u]==scc[v ...
- 静态Web开发 JQuery
伍章 JQuery 1节介绍JQuery和顶级对象 <<锋利的JQuery>>JQuery官网: http://jquery.com (下载jquery工具)JQuery在线A ...
- C# 中的装箱与拆箱
转角撞倒猪 原文 C# 中的装箱与拆箱 装箱:将一个数据项(副本)从栈中自动复制到堆中的行为. int i = 8; object o = i; // 装箱 // 首先在堆中开辟出一片区域,再将 ...
- bindService和startService的区别
区别: startService,关闭服务退出activity,service仍然处于后台运行 bindService,关闭服务退出activity直接stopService,停止服务 bindSer ...
- Linux Oracle碰到错误:ORA-27101: shared memory realm does not exist
从ITPUB上摘抄并已验证 1.实例没有启动 sqlplus /nologconnect / as sysdbastartup
- Js(Jquery)实现的弹出窗口
想实现一个弹出层,过一段时间自动消失的功能. 之前的项目中是:在页面中预先先一个<div>区域,默认隐藏或者因为没有内容不显示.当需要显示信息时,将该<div>填充上内容,并用 ...