[USACO13NOV]空荡荡的摊位Empty Stalls
题目描述
Farmer John's new barn consists of a huge circle of N stalls (2 <= N <= 3,000,000), numbered 0..N-1, with stall N-1 being adjacent to stall 0.
At the end of each day, FJ's cows arrive back at the barn one by one, each with a preferred stall they would like to occupy. However, if a cow's preferred stall is already occupied by another cow, she scans forward sequentially from this stall until she finds the first unoccupied stall, which she then claims. If she scans past stall N-1, she continues scanning from stall 0.
Given the preferred stall of each cow, please determine the smallest index of a stall that remains unoccupied after all the cows have returned to the barn. Notice that the answer to this question does not depend on the order in which the cows return to the barn.
In order to avoid issues with reading huge amounts of input, the input to this problem is specified in a concise format using K lines (1 <= K <= 10,000) each of the form:
X Y A B
One of these lines specifies the preferred stall for XY total cows: X cows prefer each of the stalls f(1) .. f(Y), where f(i) = (Ai + B) mod N. The values of A and B lie in the range 0...1,000,000,000.
Do not forget the standard memory limit of 64MB for all problems.
约翰的谷仓中有N(2 <= N <=3,000,000)个房间,编号0到N-1,这些房间排布成环状,编号0的和编号N-1的相邻。
每天傍晚,奶牛们一只一只排队回到谷仓,每头奶牛都有一个喜欢的房间,但是,如果它喜欢的房间已被其他奶牛占了,它会向前挨个探索其他房间(如果它探索过了N-1号房间,它会继续探索0号房间,以此继续下去)直到探到第一个没有被占用的房间,这时它会宣布占用这个房间。
告诉你每头奶牛喜欢的房间,当所有奶牛都找到房间后,剩下的没被占用的房间中,编号最小的是哪个。很明显,问题的答案与奶牛进入谷仓的顺序无关。
为避免输入内容过多。本题的输入数据采用一种简洁的方式:一共K(1 <= K <=10,000)行,每行格式如下:
X Y A B
表示有Y批奶牛,每批X头,也就是总共X*Y只奶牛喜欢的房间号。Y批奶牛编号1到Y,第i批X头奶牛喜欢的房间号为(A*i+B) Mod N.
A和B的取值范围为0...1,000,000,000
注意,只有64M的空间。
输入输出格式
输入格式:
Line 1: Two space-separated integers: N and K.
- Lines 2..1+K: Each line contains integers X Y A B, interpreted as
above. The total number of cows specified by all these lines will be at
most N-1. Cows can be added to the same stall by several of these
lines.
输出格式:
- Line 1: The minimum index of an unoccupied stall.
输入输出样例
说明
There are 10 stalls in the barn, numbered 0..9. The second line of input states that 3 cows prefer stall (2*1+4) mod 10 = 6, and 3 cows prefer stall (2*2+4) mod 10 = 8. The third line states that 2 cows prefer stall (0*1+1) mod 10 = 1. Line four specifies that 1 cow prefers stall (1*1+7) mod 10 = 8 (so a total of 4 cows prefer this stall).
All stalls will end up occupied except stall 5.
思路
f[i]指向向最近的一个空位;
一个类似并查集的find_father()维护;
代码
#include<cstdio>
#include<cstring>
const int maxn=3e6+;
int n,k;
int x,y,a,b,c,d;
int f[maxn];
int ff(int k){return f[k]==f[n]?k:f[k]=ff(f[k]);}
int main(){
freopen("empty.in","r",stdin);
freopen("empty.out","w",stdout);
scanf("%d%d",&n,&k);
memset(f,0x7f,sizeof(f));
for(int i=;i<=k;i++){
scanf("%d%d%d%d",&x,&y,&a,&b);
for(int i=;i<=y;i++){
c=(1ll*a*i+b)%n;
for(int j=;j<x;j++){
d=ff((c+j)%n);
f[d]=ff((d+)%n);
}
}
}
printf("%d",ff());
return ;
}
[USACO13NOV]空荡荡的摊位Empty Stalls的更多相关文章
- Luogu3090 [USACO13NOV]空荡荡的摊位Empty Stalls (动态规划)
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> ...
- 解题:USACO13NOV Empty Stalls
题面 当然可以用并查集做,不过你需要按秩合并+路径压缩(才可能过),因为数据范围十分不友好...... USACO的官方做法更为优秀.首先题目告诉我们牛们加入的前后顺序不影响结果(自己证明也很容易,显 ...
- USACO 2013 November Contest Gold 简要题解
Problem 1. Empty Stalls 扫两遍即可. Problem 2. Line of Sight 我们发现能互相看见的一对点一定能同时看见粮仓的某一段.于是转换成有n段线段,问有多少对线 ...
- BZOJ-USACO被虐记
bzoj上的usaco题目还是很好的(我被虐的很惨. 有必要总结整理一下. 1592: [Usaco2008 Feb]Making the Grade 路面修整 一开始没有想到离散化.然后离散化之后就 ...
- bzoj AC倒序
Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem ...
- bzoj1651 / P2859 [USACO06FEB]摊位预订Stall Reservations
P2859 [USACO06FEB]摊位预订Stall Reservations 维护一个按右端点从小到大的优先队列 蓝后把数据按左端点从小到大排序,顺序枚举. 每次把比右端点比枚举线段左端点小的数据 ...
- 洛谷 P3088 [USACO13NOV]挤奶牛Crowded Cows 题解
P3088 [USACO13NOV]挤奶牛Crowded Cows 题目描述 Farmer John's N cows (1 <= N <= 50,000) are grazing alo ...
- 题解 P2859 【[USACO06FEB]摊位预订Stall Reservations】
题目链接: https://www.luogu.org/problemnew/show/P2859 思路: 首先大家会想到这是典型的贪心,类似区间覆盖问题的思路,我们要将每段时间的左端点从小到大排序, ...
- [USACO06FEB]摊位预订Stall Reservations(贪心)
[USACO06FEB]摊位预订Stall Reservations 题目描述 Oh those picky N (1 <= N <= 50,000) cows! They are so ...
随机推荐
- expect下命令不能解析通配符*的问题
曾遇到这样一段代码:(Bash脚本) 1 2 3 4 5 6 7 8 9 10 11 12 #!/usr/bin/expect -f set HOST "192.168.102.1" ...
- 在js中怎样获得checkbox里选中的多个值?(jQuery)
思路:利用name属性值获取checkbox对象,然后循环判断checked属性(true表示被选中,false表示未选中).下面进行实例演示: 1.HTML结构 <input type=&qu ...
- html制作简单框架网页二 实现自己的影音驿站 操作步骤及源文件下载 (可播放mp4、avi、mpg、asx、swf各种文件的视频播放代码)
新增视频播放功能如下图: 左侧网页left.html代码如下: <meta charset="utf-8"> <body style="backgrou ...
- PL/SQL学习笔记(三)
-----创建一个序列,再创建一个表(主键是数字),通过序列生成该表的主键值. create table mytb1( m_id number primary key, m_name ) not nu ...
- Jenkins邮件扩展插件Email Extension Plugin配置使用
1.在管理插件中搜索并安装邮件扩展插件Email Extension Plugin: 2.在任务中增加构建后操作步骤,选择Editable Email Notification; 3.在高级中Add ...
- 汇编4OPCODE
opcode原理 前缀域 切换操作数大小前缀 : 066h 可以将32位的操作数切换成16位的操作数 B8 00010000 | MOV EAX,0x100 66:B8 0001 | MOV ...
- MFC (Combo-box control)下拉列表控件的使用
1.可以选择,但无法编辑状态: 选择下拉框的属性中的类型(Type)参数——Drop List; 2.如何控制Combo Box的下拉长度 1,一.在资源文件中打开对话框,选中Combo Box控件, ...
- JSP页面通过c:forEach标签循环遍历List集合
c:forEach>标签有如下属性: 属性 描述 是否必要 默认值items 要被循环的信息 否 无begin 开始的元素(0=第一个元素,1=第二个元素) 否 0end 最后一个元素(0=第一 ...
- vue-cli中添加使用less
在vue-cli中构建的项目是可以使用less的,但是查看package.json可以发现,并没有less相关的插件,所以我们需要自行安装. 第一步:安装 npm install less less- ...
- ubuntu下操作Hadoop、hdfs、hbase、zookeeper时产生的一些问题及解决办法
2019/05/29 1.在终端输入jps时,没有显示Hdfs的DataNode 在文件夹中分别找到DataNode 和Namenode的version,将Datanode的version改为与nam ...