1097 Deduplication on a Linked List (25 分)（链表）

Description

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

生词

英文	解释
deduplication	去重
duplicated	重复的
At the mean time	与此同时

题目大意：

给一个链表，去重（去掉值或者绝对值相等的），先输出删除后的链表，再输出删除了的链表。

分析：

用结构体数组存储这个链表，大小为maxn = 100000，node[i]表示地址为i的结点。在结构体中定义一个num变量，将num变量先初始化为2 * maxn。通过改变num变量的值最后sort排序来改变链表的顺序

将没有删除的结点的num标记为cnt1，cnt1为当前没有删除的结点的个数；将需要删除的结点的num标记为maxn + cnt2，cnt2表示当前删除了的结点的个数，因为一开始初始化为了2 * maxn，所以我们可以通过对num排序达到：num = 0~maxn为不删除结点，num = maxn~2maxn为删除结点，num = 2maxn为无效结点

这样sort后就会按照需要输出的顺序将结点排序，我们只需要输出前cnt1+cnt2个结点即可～
原文链接

题解

柳神的思路非常巧妙：关于num属性的用法
第一个链表的num均为0~cnt-1
第一个链表的num均为maxn~maxn+cnt2-1
其余结点的num均为2*maxn
这样就会依照num从小到大的顺序依次输出啦~

#include <bits/stdc++.h>

using namespace std;
const int maxn=100000;
struct NODE
{
    int address,key,next,num=2*maxn;
}node[maxn];
bool exist[maxn];
bool cmp(NODE a,NODE b)
{
    return a.num<b.num;
}
int main()
{
    #ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int first,n,num,cnt1=0,cnt2=0;
    scanf("%d %d",&first,&n);
    for(int i=0;i<n;i++){
        scanf("%d",&num);
        scanf("%d%d",&node[num].key,&node[num].next);
        node[num].address=num;
    }
    for(int i=first;i!=-1;i=node[i].next){
        if(exist[abs(node[i].key)]==false){
            exist[abs(node[i].key)]=true;
            node[i].num=cnt1;
            cnt1++;
        }else{
            node[i].num=maxn+cnt2;
            cnt2++;
        }
    }
    //注意这里应该是maxn而不是n，因为node的长度为maxn
    sort(node,node+maxn,cmp);
    int cnt=cnt1+cnt2;
    for(int i=0;i<cnt;i++){
        //注意这里应该是&&而不是||。
	//从离散数学的角度来看，该语句的反命题为：i==非cnt1-1 || i==非cnt2-1 
        if(i!=cnt1-1&&i!=cnt-1){
            printf("%05d %d %05d\n",node[i].address,node[i].key,node[i+1].address);
        }else
            printf("%05d %d -1\n",node[i].address,node[i].key);
    }
    return 0;
}