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. iOS - UI - UIScrollView

    1.UIScrollView 滚动视图 // 滚动视图 UIScrollView* scrollView = [[UIScrollView alloc]initWithFrame:self.view. ...

  2. (一)u-boot2013.01.01 for TQ210:《Uboot简介》

    一直想写一个s5pv210硬件平台的u-boot的移植文档,但一直都忙着没时间写.先写一些u-boot的脚本分析吧,包括makefile,mkconfig,config.mk,主要侧重于语法句意的分析 ...

  3. 来TN公司两周的记录

    刚好上两周班,现在记录一下吧. 第一周:全部是培训,关于系统的架构.BOSS.NGBOSS.编码规范.开发规范.项目文档等.听了之后对公司的技术还是有一定的了解.至少知道公司不只一个web站. 第二周 ...

  4. hg(Mercurial)版本库迁移到git版本库

    这几天没事干净搞迁移了,迁移完MVC又迁移版本库,还把工作电脑迁移了一下,开始用Win8.1了.这个迁移主要是因为实在不想在工作电脑上又装git又装hg了,点个右键出来一大堆菜单,况且现在git已经成 ...

  5. ibatis mysql replace into 返回ID

    目前只能在replace into 之后,重新查询数据

  6. codeforces 676A A. Nicholas and Permutation(水题)

    题目链接: A. Nicholas and Permutation time limit per test 1 second memory limit per test 256 megabytes i ...

  7. Java程序修改文件名

    package script; import java.io.File; import java.io.IOException; public class Realname { public stat ...

  8. 如何使用OpenShare部署和运营企业门户

    如何使用OpenShare部署和运营企业门户 这篇Blog是偏向企业内整体门户部署和运营的指南,是偏向整体管理和规划的,并不是针对终端用户的OpenShare软件操作手册,具体的操作可以上优酷看相关视 ...

  9. CSS之Generator

    这个工具可以,收藏一下.CSS Generator

  10. WIN7系统操作快捷键

    轻松访问 按住右Shift八秒钟:启用或关闭筛选键 按左 Alt+左 Shift+PrntScrn(或 PrntScrn):启用或关闭高对比度 按左 Alt+左 Shift+Num Lock :启用或 ...