Buy Tickets(线段树)
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.
#include<iostream>
#include<stdio.h>
#include<string.h>
#define N 200010
#define lson p << 1
#define rson p << 1 | 1
using namespace std; struct Nod
{
int l , r ;
int va ; // Gap
}node[N << ]; int pos[N] , val[N] , ans[N] ; void build ( int l , int r , int p )
{
node[p].l = l ;
node[p].r = r ;
node[p].va = r - l + ;
if ( l == r)
return ;
int mid = ( l + r ) >> ;
build ( l , mid , lson ) ;
build ( mid + , r , rson ) ;
} int update ( int ps , int p )
{
node[p].va-- ; //Gap - 1 ;
if ( node[p].l == node[p].r ) {
return node[p].l ; //return the position of the insert
}
if ( node[lson].va >= ps ) {
update ( ps , lson ) ;//when lson's Gap equal to or greater than the insertion location , inserted to the left
}
else {
ps -= node[lson].va ;//when Gap on the left is less than the ps , insert the right side , the right of the insertion laction is
// ps minus Gap on the left
update ( ps , rson ) ;
}
} int main ()
{
//freopen ( "a.txt" , "r" , stdin ) ;
int n ;
while (~scanf ("%d" , &n ) ) {
build ( , n , ) ;
for ( int i = ; i <= n ; i++ ) {
scanf ("%d%d" , pos + i , val + i ) ;
}
for ( int i = n ; i >= ; i-- ) {
int id = update ( pos[i] + , ) ; // get the insert position
ans[id] = val[i] ; // in ans Array
}
for ( int i = ; i < n ; i++ ) {
printf ( "%d " , ans[i] ) ;
}
printf ( "%d\n" , ans[n] ) ;
}
return ;
}
倒推法
集训时大神介绍了一种能节省空间的写法:
#include<iostream>
#include<stdio.h>
#include<string.h>
#define N 200010
#define lson o << 1 , l , mid
#define rson o << 1 | 1 , mid + 1 , r
using namespace std; int va[N << ] ; // Gap int pos[N] , val[N] , ans[N] ; void build (int o , int l , int r )
{
va[o] = r - l + ;
if ( l == r )
return ;
int mid = ( l + r ) >> ;
build ( lson ) ;
build ( rson ) ;
} int update ( int ps ,int o , int l , int r )
{
va[o]-- ; //Gap - 1 ;
if ( l == r ) {
return l ; //return the position of the insert
}
int mid = ( l + r ) >> ;
if ( va[o << ] >= ps ) {
update ( ps , lson ) ;//when lson's Gap equal to or greater than the insertion location , inserted to the left
}
else {
ps -= va[o << ] ;//when Gap on the left is less than the ps , insert the right side , the right of the insertion laction is
// ps minus Gap on the left
update ( ps , rson ) ;
}
} int main ()
{
//freopen ( "a.txt" , "r" , stdin ) ;
int n ;
while (~scanf ("%d" , &n ) ) {
build ( , , n ) ;
for ( int i = ; i <= n ; i++ ) {
scanf ("%d%d" , pos + i , val + i ) ;
}
for ( int i = n ; i >= ; i-- ) {
int id = update ( pos[i] + , , , n ) ; // get the insert position
ans[id] = val[i] ; // in ans Array
}
for ( int i = ; i < n ; i++ ) {
printf ( "%d " , ans[i] ) ;
}
printf ( "%d\n" , ans[n] ) ;
}
return ;
}
省了4000k
Buy Tickets(线段树)的更多相关文章
- [poj2828] Buy Tickets (线段树)
线段树 Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must ...
- 【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(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 倒序插入 节点空位预留(思路巧妙)
Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 19725 Accepted: 9756 Desc ...
- Buy Tickets(线段树)
Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 16607 Accepted: 8275 Desc ...
- 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 or 树状数组+二分)
题目链接:http://poj.org/problem?id=2828 题意就是给你n个人,然后每个人按顺序插队,问你最终的顺序是怎么样的. 反过来做就很容易了,从最后一个人开始推,最后一个人位置很容 ...
- poj2828 Buy Tickets (线段树 插队问题)
Buy Tickets Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 22097 Accepted: 10834 Des ...
- POJ 2828 Buy Tickets | 线段树的喵用
题意: 给你n次插队操作,每次两个数,pos,w,意为在pos后插入一个权值为w的数; 最后输出1~n的权值 题解: 首先可以发现,最后一次插入的位置是准确的位置 所以这个就变成了若干个子问题, 所以 ...
- POJ 2828 Buy Tickets(线段树·插队)
题意 n个人排队 每一个人都有个属性值 依次输入n个pos[i] val[i] 表示第i个人直接插到当前第pos[i]个人后面 他的属性值为val[i] 要求最后依次输出队中各个人的属性 ...
随机推荐
- 『随笔』WCF开发那些需要注意的坑
执行如下 批处理:"C:\Program Files\Microsoft SDKs\Windows\v6.0A\Bin\svcutil.exe" http://127.0.0.1: ...
- Ajax在调用含有SoapHeader的webservice方法
· [WebService(Namespace = "http://tempuri.org/")] · [WebServiceBinding ...
- [USACO]6.1.3 cow xor(二进制+Trie)
题意:给你一个序列(n<=100000),求出一个连续的子序列[i,j]使得ai xor ai+1 xor…… xor aj最大,求出这个最大值(其中每个数<=2^21) 分析:题目和求一 ...
- 转-JS中document对象详解
对象属性 document.title //设置文档标题等价于HTML的<title>标签 document.bgColor //设置页面背景色 document.fgColor //设置 ...
- Sublime-jQueryDocs
Package Control Messages======================== jQueryDocs---------- This package shows a selected ...
- Linux大文件已删除,但df查看已使用的空间并未减少解决
在我的生活当中遇到磁盘快满了,这时候准备去删除一些大文件 于是我使用ncdu 查看了一下当前系统占用资源比较多的是那些文件,结果一看是elasticsearch的日志文件,好吧,竟然找到源头了,那就把 ...
- MVC4学习笔记(一)
1.查询 1)Controllers /// <summary> /// 数据上下文对象 /// </summary> OumindBlogEntities db = new ...
- 【UVA 401】BUPT 2015 newbie practice #2 div2-B-Palindromes
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102419#problem/B A regular palindrome is a str ...
- SourceTree&Git部分名词解释
SourceTree&Git部分名词解释 克隆(clone):从远程仓库URL加载创建一个与远程仓库一样的本地仓库 提交(commit):将暂存文件上传到本地仓库(我们在Finder中对本地仓 ...
- Memory Allocation API In Linux Kernel && Linux Userspace、kmalloc vmalloc Difference、Kernel Large Section Memory Allocation
目录 . 内核态(ring0)内存申请和用户态(ring3)内存申请 . 内核态(ring0)内存申请:kmalloc/kfree.vmalloc/vfree . 用户态(ring3)内存申请:mal ...