Buy Tickets

Time Limit: 1 Sec  Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=2795

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 ≤ iN). 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

题意

给你一个队列,有一群人来插队,然后让你输出最后队列的样子

题解:

感觉是个很傻B的题,但是为什么我非得用线段树来做呢?= =

vector亲测被T掉

代码:

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#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 200050
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/*
;
inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH[Num--]+48);
puts("");
}
*/
//**************************************************************************************
inline ll read()
{
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
} struct node
{
int l,r,v;
}; node a[maxn*]; void build(int x,int l,int r)
{
a[x].l=l,a[x].r=r;
a[x].v=r-l+;
if(l==r)
return;
int mid=(l+r)>>;
build(x<<,l,mid);
build(x<<|,mid+,r);
} int update(int x,int val)
{
int l=a[x].l,r=a[x].r;
if(l==r)
{
a[x].v=;
return l;
}
else
{
int pos=;
if(a[x<<].v>=val)
pos=update(x<<,val);
else
pos=update(x<<|,val-a[x<<].v);
a[x].v=a[x<<].v+a[x<<|].v;
return pos;
}
} int n;
int b[maxn];
int c[maxn];
int pos[maxn];
int main()
{
while(scanf("%d",&n)!=EOF)
{
memset(a,,sizeof(a));
build(,,n);
for(int i=;i<n;i++)
b[i]=read(),c[i]=read();
for(int i=n-;i>=;i--)
pos[update(,b[i]+)]=c[i];
for(int i=;i<=n;i++)
printf("%d ",pos[i]);
puts("");
}
}

poj 2828 buy Tickets 用线段树模拟带插入的队列的更多相关文章

  1. poj 2828 Buy Tickets 【线段树点更新】

    题目:id=2828" target="_blank">poj 2828 Buy Tickets 题意:有n个人排队,每一个人有一个价值和要插的位置,然后当要插的位 ...

  2. POJ 2828 Buy Tickets (线段树 单点更新 变形)

    题目链接 题意:有N个人排队,给出各个人想插队的位置和标识,要求输出最后的序列. 分析:因为之前的序列会因为插队而变化,如果直接算时间复杂度很高,所以可以用 线段树逆序插入,把序列都插到最后一层,le ...

  3. 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 ...

  4. POJ - 2828 Buy Tickets(线段树单点更新)

    http://poj.org/problem?id=2828 题意 排队买票,依次给出当前人要插队的位置,每个人有个编号,然后问你最后整个的序列是什么? 分析 最后一个人的要插入的位置是确定的,所以逆 ...

  5. 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 ...

  6. poj 2828 Buy Tickets (线段树 单节点 查询位置更新)

    Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 15533   Accepted: 7759 Desc ...

  7. POJ 2828 Buy Tickets (线段树 || 树状数组)

    题目大意 一些小朋友在排队,每次来一个人,第i个人会插到第x个人的后面.权值为y.保证x∈[0,i-1]. 按照最后的队伍顺序,依次输出每个人的权值. 解题分析 好气吖.本来是在做splay练习,然后 ...

  8. poj 2828 Buy Tickets【线段树 单点更新】

    倒着插,先不理解意思,后来看一篇题解说模拟一下 手动模拟一下就好理解了----- 不过话说一直写挫---一直改啊----- 好心塞------ #include <cstdio> #inc ...

  9. 【POJ】2828 Buy Tickets(线段树+特殊的技巧/splay)

    http://poj.org/problem?id=2828 一开始敲了个splay,直接模拟. tle了.. 常数太大.. 好吧,说是用线段树.. 而且思想很拽.. (貌似很久以前写过貌似的,,) ...

随机推荐

  1. 一个爬取https和http通用的工具类(JDK自带的URL的用法)

    今天在java爬取天猫的时候因为ssl报错,所以从网上找了一个可以爬取https和http通用的工具类.但是有的时候此工具类爬到的数据不全,此处不得不说python爬虫很厉害. package cn. ...

  2. 动态规划_01背包问题_Java实现

    原文地址:http://blog.csdn.net/ljmingcom304/article/details/50328141 本文出自:[梁敬明的博客] 1.动态规划 什么是动态规划?动态规划就是将 ...

  3. python设计模式之装饰器详解(三)

    python的装饰器使用是python语言一个非常重要的部分,装饰器是程序设计模式中装饰模式的具体化,python提供了特殊的语法糖可以非常方便的实现装饰模式. 系列文章 python设计模式之单例模 ...

  4. 日常开发技巧:x11-forward,使用远程机器的gui程序

    背景 日常用过ssh登录服务器进行工作,尽管大部分时间,都只需要终端操作,编辑源码也是vim就够用了. 但有时候,还是需要使用gui程序的,比如打开一份pdf,word,ppt,excel等. 碰到这 ...

  5. ubuntu sougou输入法

    1, 打开搜狗输入法Linux版的官网http://pinyin.sogou.com/linux/?r=pinyin,并下载你需要的版本,这里选择64位版. 2,在Ubuntu14.01下可以直接点击 ...

  6. https、socket、http协议

    一.https https 其实是由两部分组成:http+ssl(Secure Sockets Layer 安全套接层)/tls(Transport Layer Security 继任者安全传输层), ...

  7. u-boot界面添加命令[demo]

    目标板:2440 如何在u-boot界面中增加命令 在/common/目录下建立文件,调用执行函数do_bootm就行,然后在修改Makefile,就OK了. 比如在u-boot界面添加命令test ...

  8. 7.Python3标准库--文件系统

    ''' Python的标准库中包含大量工具,可以处理文件系统中的文件,构造和解析文件名,还可以检查文件内容. 处理文件的第一步是要确定处理的文件的名字.Python将文件名表示为简单的字符串,另外还提 ...

  9. Java集合里的一些“坑”

    这里主要谈下Java集合在使用中容易被忽略.又容易出现的两个“坑”,一个是集合与数组互相转换,另一个是集合遍历删除.主要通过代码演示. 一.集合与数组互相转换中的“坑” //Test1.java pa ...

  10. vmware linux虚拟机连接ip设置

    首先: 点击VMware 编辑->虚拟网络编辑器: 然后选中VMnet8的查看NAT设置: 上图第二步(记下红框中网关地址和子网掩码): 第三步(用于设置虚拟机地址范围): 接下来就是设置虚拟机 ...