B. Queue
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

During the lunch break all n Berland State University students lined up in the food court. However, it turned out that the food court, too, has a lunch break and it temporarily stopped working.

Standing in a queue that isn't being served is so boring! So, each of the students wrote down the number of the student ID of the student that stands in line directly in front of him, and the student that stands in line directly behind him. If no one stands before or after a student (that is, he is the first one or the last one), then he writes down number 0 instead (in Berland State University student IDs are numerated from 1).

After that, all the students went about their business. When they returned, they found out that restoring the queue is not such an easy task.

Help the students to restore the state of the queue by the numbers of the student ID's of their neighbors in the queue.

Input

The first line contains integer n (2 ≤ n ≤ 2·105) — the number of students in the queue.

Then n lines follow, i-th line contains the pair of integers ai, bi (0 ≤ ai, bi ≤ 106), where ai is the ID number of a person in front of a student and bi is the ID number of a person behind a student. The lines are given in the arbitrary order. Value 0 is given instead of a neighbor's ID number if the neighbor doesn't exist.

The ID numbers of all students are distinct. It is guaranteed that the records correspond too the queue where all the students stand in some order.

Output

Print a sequence of n integers x1, x2, ..., xn — the sequence of ID numbers of all the students in the order they go in the queue from the first student to the last one.

Examples
input
4
92 31
0 7
31 0
7 141
output
92 7 31 141 
Note

The picture illustrates the queue for the first sample.

题意:告诉你前驱和后继,求原序列;

思路:两个数位置相差2;

   找到前后的起点,两个方向扫一遍;

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define esp 0.00000000001
const int N=3e5+,M=1e6+,inf=1e9,mod=1e9+;
int a[M];
int b[M];
int ans[M];
int flag[M]; int main()
{
int x,y,z,i,t;
scanf("%d",&x);
for(i=;i<x;i++)
{
scanf("%d%d",&y,&z);
a[y]=z;
b[z]=y;
flag[y]++;
flag[z]++;
}
int st=;
int ji=;
while(a[st])
{
ans[ji+]=a[st];
flag[a[st]]=;
st=a[st];
ji+=;
}
if(x&)
{
int en=;
for(i=;i<=M;i++)
if(flag[i]&&!a[i])
en=i;
int ji=x;
while(en)
{
ans[ji]=en;
en=b[en];
ji-=;
}
}
else
{
int ji=x-;
int en=;
while(b[en])
{
ans[ji]=b[en];
en=b[en];
ji-=;
}
}
for(i=;i<=x;i++)
printf("%d ",ans[i]);
return ;
}
B. Queue
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

During the lunch break all n Berland State University students lined up in the food court. However, it turned out that the food court, too, has a lunch break and it temporarily stopped working.

Standing in a queue that isn't being served is so boring! So, each of the students wrote down the number of the student ID of the student that stands in line directly in front of him, and the student that stands in line directly behind him. If no one stands before or after a student (that is, he is the first one or the last one), then he writes down number 0 instead (in Berland State University student IDs are numerated from 1).

After that, all the students went about their business. When they returned, they found out that restoring the queue is not such an easy task.

Help the students to restore the state of the queue by the numbers of the student ID's of their neighbors in the queue.

Input

The first line contains integer n (2 ≤ n ≤ 2·105) — the number of students in the queue.

Then n lines follow, i-th line contains the pair of integers ai, bi (0 ≤ ai, bi ≤ 106), where ai is the ID number of a person in front of a student and bi is the ID number of a person behind a student. The lines are given in the arbitrary order. Value 0 is given instead of a neighbor's ID number if the neighbor doesn't exist.

The ID numbers of all students are distinct. It is guaranteed that the records correspond too the queue where all the students stand in some order.

Output

Print a sequence of n integers x1, x2, ..., xn — the sequence of ID numbers of all the students in the order they go in the queue from the first student to the last one.

Examples
input
4
92 31
0 7
31 0
7 141
output
92 7 31 141 
Note

The picture illustrates the queue for the first sample.

Codeforces Round #279 (Div. 2) B. Queue的更多相关文章

  1. Codeforces Round #279 (Div. 2)B. Queue(构造法,数组下标的巧用)

    这道题不错,思维上不难想到规律,但是如何写出优雅的代码比较考功力. 首先第一个人的序号可以确定,那么接下来所有奇数位的序号就可以一个连一个的确定了.然后a[i].first==0时的a[i].seco ...

  2. Codeforces Round #279 (Div. 2) B - Queue 水题

    #include<iostream> #include<mem.h> using namespace std; ],q[]; int main() { int n,x,y; m ...

  3. Codeforces Round #279 (Div. 2) ABCDE

    Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/outpu ...

  4. 水题 Codeforces Round #303 (Div. 2) D. Queue

    题目传送门 /* 比C还水... */ #include <cstdio> #include <algorithm> #include <cstring> #inc ...

  5. 【Codeforces Round#279 Div.2】B. Queue

    这题看别人的.就是那么诚实.http://www.cnblogs.com/zhyfzy/p/4117481.html B. Queue During the lunch break all n Ber ...

  6. Codeforces Round #279 (Div. 2) 题解集合

    终于有场正常时间的比赛了...毛子换冬令时还正是好啊233 做了ABCD,E WA了3次最后没搞定,F不会= = 那就来说说做的题目吧= = A. Team Olympiad 水题嘛= = 就是个贪心 ...

  7. Codeforces Round #303 (Div. 2) D. Queue 傻逼题

    C. Woodcutters Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/545/probl ...

  8. Codeforces Round #279 (Div. 2) C. Hacking Cypher 机智的前缀和处理

    #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #incl ...

  9. Codeforces Round #303 (Div. 2) D. Queue —— 贪心

    题目链接:http://codeforces.com/problemset/problem/545/D 题解: 问经过调整,最多能使多少个人满意. 首先是排序,然后策略是:如果这个人对等待时间满意,则 ...

随机推荐

  1. Python慢,为啥还有大公司用?

    PyCon 是全世界最大的以 Python 编程语言 为主题的技术大会,大会由 Python 社区组织,每年举办一次.在 Python 2017 上,Instagram 的工程师们带来了一个有关 Py ...

  2. 搭建高性能Jboss负载均衡集群

    版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/a1314517love/article/details/26836357 负载均衡集群是由两台或者两 ...

  3. mysql 数据库查询最后两条数据

    版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/u011925175/article/details/24186917  有一个mysql数据库的 ...

  4. 爬虫——请求库之requests

    阅读目录 一 介绍 二 基于GET请求 三 基于POST请求 四 响应Response 五 高级用法 一 介绍 #介绍:使用requests可以模拟浏览器的请求,比起之前用到的urllib,reque ...

  5. flask—信号(blinker)

    Flask框架中的信号基于blinker,主要是让开发者在flask请求过程中定制一些用户行为. 安装blinker pip3 install blinker 1.内置信号 request_start ...

  6. centos7命令2

    都代表一个程序find / -name '*name*' 查找根目录下的包含name的文件名 权限不足的话在命令前面加上sudo find / -size 1G 查找根目录下文件大小为1G的文件 fi ...

  7. Python3 计算城市距离

    利用上一篇得到的城市经纬度算城市距离 import requests from math import radians, cos, sin, asin, sqrt def geocode(addres ...

  8. Kattis - bank 【简单DP】

    Kattis - bank [简单DP] Description Oliver is a manager of a bank near KTH and wants to close soon. The ...

  9. splay伸展树模板

      普通版本: struct SplayTree { ; ], key[maxn], val[maxn], sz[maxn], lz[maxn], fa[maxn]; , ) { ch[x][]=ch ...

  10. Python学习示例源码

    函数和函数式编程 函数定义: 函数调用: 过程定义: 过程调用: 面向过程的编程方法: """ 面向对象-----类------class 面向过程-----过程---- ...