Buy Tickets
Buy Tickets
Time Limit: 4000MS Memory Limit: 65536K
Total Submissions: 16010 Accepted: 7983
Description
Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue…
The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. Now, he had to travel by train to Mianyang, Sichuan Province for the winter camp selection of the national team of Olympiad in Informatics.
It was one o’clock a.m. and dark outside. Chill wind from the northwest did not scare off the people in the queue. The cold night gave the Little Cat a shiver. Why not find a problem to think about? That was none the less better than freezing to death!
People kept jumping the queue. Since it was too dark around, such moves would not be discovered even by the people adjacent to the queue-jumpers. “If every person in the queue is assigned an integral value and all the information about those who have jumped the queue and where they stand after queue-jumping is given, can I find out the final order of people in the queue?” Thought the Little Cat.
Input
There will be several test cases in the input. Each test case consists of N + 1 lines where N (1 ≤ N ≤ 200,000) is given in the first line of the test case. The next N lines contain the pairs of values Posi and Vali in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Posi and Vali are as follows:
Posi ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Posi-th person in the queue. The booking office was considered the 0th person and the person at the front of the queue was considered the first person in the queue.
Vali ∈ [0, 32767] — The i-th person was assigned the value Vali.
There no blank lines between test cases. Proceed to the end of input.
Output
For each test cases, output a single line of space-separated integers which are the values of people in the order they stand in the queue.
Sample Input
4
0 77
1 51
1 33
2 69
4
0 20523
1 19243
1 3890
0 31492
Sample Output
77 33 69 51
31492 20523 3890 19243
Hint
The figure below shows how the Little Cat found out the final order of people in the queue described in the first test case of the sample input.
Source
POJ Monthly–2006.05.28, Zhu, Zeyuan
插队问题,就是每个人都有自己的位置,当自己的位置有人时,那个人就要向后移动,输出对应位置上人的vail,所以对于最后的人,他的位置是固定的,而之前的人的位置是变动的,所以可以从后向前,进行插入.
但我发现一个奇怪的问题,g++超时,c++ AC,将代码的输出改成每个数据后面都有空格g++就可以AC 神题
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#define LL long long
using namespace std;
const int MAX = 210000;
int num[MAX*4];//记录num[i]对应的区间还可以插入的位置个数
int w[MAX];
int a[MAX],ww[MAX];
void Build(int L,int R,int site)//初始化
{
if(L==R)
{
num[site]=1;
return ;
}
int mid = (L+R)>>1;
Build(L,mid,site<<1);
Build(mid+1,R,site<<1|1);
num[site]=num[site<<1]+num[site<<1|1];
}
void update(int L,int R,int s,int W,int site)
{
if(L==R)
{
num[site]=0;
w[L]=W;
return ;
}
int mid=(L+R)>>1;
if(num[site<<1]>=s)//如果左子树的位置的个数比它要插入的位置大说明可以插入
{
update(L,mid,s,W,site<<1);
}
else//如果能插入左子树,则插入右子树中,相对右子树它插入的位置为(s-num[site<<1]);
{
update(mid+1,R,s-num[site<<1],W,site<<1|1);
}
num[site]=num[site<<1]+num[site<<1|1];
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
{
scanf("%d %d",&a[i],&ww[i]);
}
Build(1,n,1);
for(int i=n;i>=1;i--)//逆序插入
{
update(1,n,a[i]+1,ww[i],1);
}
for(int i=1;i<=n;i++)//输出
{
if(i!=1)
{
printf(" ");
}
printf("%d",w[i]);
}
printf("\n");
}
return 0;
}
Buy Tickets的更多相关文章
- POJ2828 Buy Tickets[树状数组第k小值 倒序]
Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 19012 Accepted: 9442 Desc ...
- POJ 2828 Buy Tickets(线段树 树状数组/单点更新)
题目链接: 传送门 Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Description Railway tickets were d ...
- Buy Tickets(线段树)
Buy Tickets Time Limit:4000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- 【poj2828】Buy Tickets 线段树 插队问题
[poj2828]Buy Tickets Description Railway tickets were difficult to buy around the Lunar New Year in ...
- poj-----(2828)Buy Tickets(线段树单点更新)
Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 12930 Accepted: 6412 Desc ...
- poj 2828 Buy Tickets (线段树(排队插入后输出序列))
http://poj.org/problem?id=2828 Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Total Submissio ...
- poj 2828 Buy Tickets【线段树单点更新】【逆序输入】
Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 16273 Accepted: 8098 Desc ...
- 线段树(倒序操作):POJ 2828 Buy Tickets
Buy Tickets Description Railway tickets were difficult to buy around the Lunar New Year in China, ...
- Buy Tickets(线段树)
Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 16607 Accepted: 8275 Desc ...
随机推荐
- HashedWheelTimer
HashedWheelTimer 是根据 Hashed and Hierarchical Timing Wheels: Data Structuresfor the Efficient Impleme ...
- Java Servlet(二):servlet配置及生命周期相关(jdk7+tomcat7+eclipse)
该篇文章记录了Servlet配置相关用法及Servlet在Servlet容器中生命周期方法. Tomcat是一个Servlet容器: 1.Servlet容器管理了Servlet的整个生命周期,并调用s ...
- 重复点击主界面(TabBar)按钮刷新界面--点击状态栏回到顶部
1.监听按钮点击 2.判断是否是点击的同一个按钮(记录上次点击的按钮) 3.当重复点击相同按钮时,需要获取当前按钮对应控制器刷新界面 3.1 判断是否重复点击按钮,代码写在哪里? ...
- linux第1天 fork exec 守护进程
概念方面 文件是对I/O设备的抽象表示.虚拟存储器是对主存和磁盘I/O设备的抽象表示.进程则是对处理器.主存和I/O设备的抽象表示 中断 早期是没有进程这个概念,当出现中断技术以后才出现进程这个概念 ...
- 前端开发与Seo基础
网页代码优化 1:<title>标题:强调重点,重点关键词放在前面,每个页面的title尽量不相同 2:<meta keywords>关键词:列举出几个重要关键词, ...
- C# 实现 单例模式
http://blog.sina.com.cn/s/blog_75247c770100yxpb.html
- Logic and Fault simulation
fault simulation是指对fault circuit的simulation,来locate manufacturing defects并且进行fault diagnosis. logic ...
- Power Gating的设计(模块)
Switching Fabric的设计: 三种架构:P沟道的switch vdd(header switch),N沟道的switch vss(footer switch),两个switch. 但是如果 ...
- Centos安装wine等组件的问题
linux下安装wine可以从源码编译安装,但一般都觉得麻烦,所以尽量利用yum进行安装,解决很多包的依赖关系. 首先安装一个epelrpm -ivh http://dl.fedoraproject. ...
- Java笔试题解答和部分面试题
面试类 银行类的问题 问题一:在多线程环境中使用HashMap会有什么问题?在什么情况下使用get()方法会产生无限循环? HashMap本身没有什么问题,有没有问题取决于你是如何使用它的.比如,你 ...