Buy Tickets
Time Limit: 4000MS   Memory Limit: 65536K
Total Submissions: 16607   Accepted: 8275

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

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.

   

题解:

刚开始没理解题意,链表错了,

题意:

有n个人在火车站买票,由于天黑,所以你插队没人会看见,现在给出n个人的插队目标(允许自己前面有几个人),和他的价值(在这里没有用,只是在输出时用),让你求出n个人目标完成后,输出他们所对应的价值。

因为当允许前面插得人一样的时候,后面的会查到前面;所以从后往前;

代码:

/*#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int INF=0x3f3f3f3f;
typedef long long LL;
const int MAXN= 200010;
int next[MAXN];
int m[MAXN];
int main(){
int N,cur;
while(~scanf("%d",&N)){
next[0]=0;mem(next,0);
for(int i=1;i<=N;i++){
scanf("%d%d",&cur,m+i);
next[i]=next[cur];
next[cur]=i;
}
for(int i=next[0];i!=0;i=next[i]){
printf("%d ",m[i]);
}puts("");
}
return 0;
}*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define L tree[root].l
#define R tree[root].r
#define V tree[root].v
#define NUM tree[root].num
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define NOW NUM=tree[root<<1].num+tree[root<<1|1].num;
const int MAXN=200010;
struct Node{
int l,r,num;
};
Node tree[1000000];
const int INF=0x3f3f3f3f;
typedef long long LL;
int ans[MAXN],a[MAXN],b[MAXN],k;
void made(int root,int l,int r){
L=l;R=r;
NUM=(r-l+1);
if(l==r)return;
int mid=(l+r)>>1;
made(lson);
made(rson);
}
void update(int root,int v,int pos){
if(L==R){
NUM=0;
ans[L]=v;
return;
}
else{
if(tree[root<<1].num>=pos)update(root<<1,v,pos);
else update(root<<1|1,v,pos-tree[root<<1].num);//应该减去左树的num。。。
}
NOW;
}
int main(){
int N;
while(~scanf("%d",&N)){
made(1,1,N);
for(int i=0;i<N;i++){
scanf("%d%d",&a[i],&b[i]);
}
for(int i=N-1;i>=0;i--)update(1,b[i],a[i]+1);//加1
for(int i=1;i<N;i++)printf("%d ",ans[i]);
printf("%d\n",ans[N]);
}
return 0;
}

Buy Tickets(线段树)的更多相关文章

  1. [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 ...

  2. 【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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. poj2828 Buy Tickets (线段树 插队问题)

    Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 22097   Accepted: 10834 Des ...

  8. 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 ...

  9. POJ 2828 Buy Tickets(线段树&#183;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

  1. ASP.NET页面之间数据传递的几种方法

    1)Request.QueryString   在ASP时代,这个是较常用的方法,到了ASP.NET,好像用的人不多了,但是不管怎么说,这是一个没有过时,且很值得推荐的方法,因为不管是ASP还是ASP ...

  2. Android四大组件之Activity详解

    一.Activity的概要说明 我看过Activity的源码,Activity类注释大概是这样解释的:几乎所有的Activity都是与用户交互用的,我想用了一个几乎的意思应该是排除一些纯展示界面吧,因 ...

  3. AOP annotation

    1.xml文件 <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http ...

  4. 创建txt格式文本日志

    公共方法(可以将其放到类库里边): #region 记录日志 #region 写日志 /// <summary> /// 写日志 /// </summary> /// < ...

  5. Githut Token (hidden): Githut 安装验证

    登录https://github.com 进入https://github.com/settings/profile 参考 http://jingyan.baidu.com/article/22fe7 ...

  6. Word2007中如何插入参考文献

    很多国内的期刊杂志都只能使用word模板,导致插入参考文献成了件麻烦事,这时特别怀念Latex的便捷.于是找到一篇介绍word2007里插入参考文献的好方法,就是利用尾注的方法使文章的参考文献标号可以 ...

  7. css为网页顶部和底部都加入背景图

    网页背景图是我们常用的功能,一般来说.给网页加一个背景图,只要在网页的body标签中加入css属性就行. 代码如下:<body style="background-image:url( ...

  8. SumoLogic

    SumoLogic>>>Loggly. https://diyunpeng.loggly.com/setup MonitorWare http://www.monitorware.c ...

  9. 在C#中调用API获取网络信息和流量

    原文 在C#中调用API获取网络信息和流量 最近一项目中要求显示网络流量,而且必须使用C#. 事实上,调用 IpHlpApi.dll 的 GetIfTable API 可以轻易获得网络信息和网络流量. ...

  10. Docker学习总结之Run命令介绍

    Docker学习总结之Run命令介绍 本文由Vikings(http://www.cnblogs.com/vikings-blog/) 原创,转载请标明.谢谢! 在使用Docker时,执行最多的命令某 ...