PAT_A1133#Splitting A Linked List

Source:

PAT A1133 Splitting A Linked List (25 分）

Description:

Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

Input Specification:

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

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

Address Data Next

where Address is the position of the node, Data is an integer in [, and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218

Sample Output:

33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1

Keys:

• 链表
• 散列（Hash）

Code:

 /*
 Data: 2019-08-09 14:34:04
 Problem: PAT_A1133#Splitting A Linked List
 AC: 23:34

 题目大意：
 重排单链表，不改变原先顺序的前提下，使得
 1.负数在前，正数在后；
 2.正数中，小于K元素在前，大于K的在后；
 输入：
 第一行给出，首元素地址，结点数N<=1e5，阈值K
 接下来N行，地址，数值，下一个地址
 输出：
 地址，数值，下一个地址

 基本思路：
 结构体存储链表信息
 按顺序存储链表中有效的数值，再按要求排序，输出；
 */
 #include<cstdio>
 #include<vector>
 using namespace std;
 const int M=1e5+;
 struct node
 {
     int adr,data,nxt;
 }link[M],t;
 vector<node> p1,p2,p;

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     int fst,n,k;
     scanf("%d%d%d", &fst,&n,&k);
     for(int i=; i<n; i++)
     {
         scanf("%d%d%d", &t.adr,&t.data,&t.nxt);
         link[t.adr] = t;
     }
     while(fst != -)
     {
         if(link[fst].data < )
             p.push_back(link[fst]);
         else if(link[fst].data <=k)
             p1.push_back(link[fst]);
         else
             p2.push_back(link[fst]);
         fst = link[fst].nxt;
     }
     p.insert(p.end(),p1.begin(),p1.end());
     p.insert(p.end(),p2.begin(),p2.end());

     for(int i=; i<p.size(); i++)
     {
         if(i!=)    printf("%05d\n", p[i].adr);
         printf("%05d %d ", p[i].adr, p[i].data);
     }
     printf("-1");

     return ;
 }