Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, given L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (<= 105) which is the total number of nodes. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Key Next

where Address is the position of the node, Key is an integer of which absolute value is no more than 104, and Next is the position of the next node.

Output Specification:

For each case, output the resulting linked list first, then the removed list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 5
99999 -7 87654
23854 -15 00000
87654 15 -1
00000 -15 99999
00100 21 23854

Sample Output:

00100 21 23854
23854 -15 99999
99999 -7 -1
00000 -15 87654
87654 15 -1
 #include<stdio.h>
#include<math.h>
#include<set>
#include<algorithm>
#include<vector>
using namespace std;
struct node
{
int val;
int next,id;
};
node Link[];
int main()
{
int head,n,id,val,next;
scanf("%d%d",&head,&n);
for(int i = ;i < n ;++i)
{
scanf("%d%d%d",&id,&val,&next);
Link[id].val = val;
Link[id].next = next;
Link[id].id = id;
}
int p = head;
set<int> ss;
vector<node> vv,vv2;
while(p != -)
{
int tem = abs(Link[p].val);
if(ss.count(tem) == )
{
ss.insert(tem);
vv.push_back(Link[p]);
}
else vv2.push_back(Link[p]);
p = Link[p].next;
}
p = head;
for(int i = ; i < (int)vv.size() - ;++i)
{
printf("%05d %d %05d\n",vv[i].id,vv[i].val,vv[i+].id);
}
if(vv.size() > )
printf("%05d %d -1\n",vv[vv.size()-].id,vv[vv.size()-].val); for(int i = ;i < (int)vv2.size() -;++i)
{
printf("%05d %d %05d\n",vv2[i].id,vv2[i].val,vv2[i+].id);
}
if(vv2.size() > )
printf("%05d %d -1\n",vv2[vv2.size()-].id,vv2[vv2.size()-].val);
return ;
}

1097. Deduplication on a Linked List (25)的更多相关文章

  1. PAT (Advanced Level) Practise - 1097. Deduplication on a Linked List (25)

    http://www.patest.cn/contests/pat-a-practise/1097 Given a singly linked list L with integer keys, yo ...

  2. PAT Advanced 1097 Deduplication on a Linked List (25) [链表]

    题目 Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplica ...

  3. PAT (Advanced Level) 1097. Deduplication on a Linked List (25)

    简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...

  4. PAT甲级题解-1097. Deduplication on a Linked List (25)-链表的删除操作

    给定一个链表,你需要删除那些绝对值相同的节点,对于每个绝对值K,仅保留第一个出现的节点.删除的节点会保留在另一条链表上.简单来说就是去重,去掉绝对值相同的那些.先输出删除后的链表,再输出删除了的链表. ...

  5. 【PAT甲级】1097 Deduplication on a Linked List (25 分)

    题意: 输入一个地址和一个正整数N(<=100000),接着输入N行每行包括一个五位数的地址和一个结点的值以及下一个结点的地址.输出除去具有相同绝对值的结点的链表以及被除去的链表(由被除去的结点 ...

  6. PAT甲级——1097 Deduplication on a Linked List (链表)

    本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/91157982 1097 Deduplication on a L ...

  7. pat1097. Deduplication on a Linked List (25)

    1097. Deduplication on a Linked List (25) 时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 ...

  8. PAT 1097 Deduplication on a Linked List[比较]

    1097 Deduplication on a Linked List(25 分) Given a singly linked list L with integer keys, you are su ...

  9. PAT 1097. Deduplication on a Linked List (链表)

    Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated ...

随机推荐

  1. maven 本地仓库nexus的安装

    首先我们将nexus下载下来:http://www.sonatype.org/downloads/nexus-latest-bundle.zip 下载下来之后我们将文件解压,解压完成之后,我们首先,打 ...

  2. 【安卓面试题】Activity和Task的启动模式有哪些?每种含义是什么?举例说明各自的应用场景

    Activity和Task的启动模式有哪些?每种含义是什么?举例说明各自的应用场景 Activity的启动模式 (Launchmode) 有4种 1.standard 默认模式,不需要配置 含义: 启 ...

  3. hdu 3062 2-SAT问题

    思路:裸的2-SAT. #include<map> #include<set> #include<cmath> #include<queue> #inc ...

  4. 初识 Asp.Net内置对象之Session对象

    Session对象 Session对象用于存储在多个页面调用之间特定用户的信息.Session对象只针对单一网站使用者,不同的客户端无法相互访问.Session对象中止联机机器离现时,,也就是当网站使 ...

  5. C# 序列化JavaScriptSerializer

    1.首先引入 System.Web.Extensions.dll 2.写入命名空间 System.Web.Script.Serialization 3.实现序列化. class Program { s ...

  6. HTML5游戏开发,剪刀石头布小游戏案例

    剪刀石头布,非常可爱的小游戏,相信大家都非常的怀念这款小游戏,小时候也玩过很多次,陪伴着我的童年的成长,现在是不是还会玩一下,剪刀石头布游戏的规则我们都知道是:剪刀剪布,石头砸剪刀,布包石头.跟朋友. ...

  7. 手把手教你认识并搭建Nginx

    手把手教你认识并搭建Nginx Nginx (“engine x”) 是一个高性能的 HTTP 和 反向代理 服务器,也是一个 IMAP/POP3/SMTP 代理服务器. Nginx 是由 Igor ...

  8. 【Cocos2d入门教程三】HelloWorld之一目了然

    什么程序都是从HelloWorld先开始.同样Cocos2d-x我们先从HelloWorld进行下手.下面是HelloWorld的运行完成图: 建立好的Cocos游戏项目中会有两个比较常用接触的文件夹 ...

  9. 集合类学习之HashMap

    一.HashMap概述 HashMap基于哈希表的 Map 接口的实现.此实现提供所有可选的映射操作,并允许使用 null 值和 null 键.(除了不同步和允许使用 null 之外,HashMap  ...

  10. C#创建Windows服务入门图解(VS2010)

    C#创建Windows服务入门图解(VS2010) Windows服务大家都知道,比如Audio.Theme都是大家比较熟悉的服务,他们可以设为自动启动的,并且在注册表的开机自启动项里是没有痕迹的.所 ...