02-线性结构3 Reversing Linked List

02-线性结构3 Reversing Linked List   (25分)

  • 时间限制：400ms
  • 内存限制：64MB
  • 代码长度限制：16kB
  • 判题程序：系统默认
  • 作者：陈越
  • 单位：浙江大学

https://pta.patest.cn/pta/test/3512/exam/4/question/62614

Given a constant KKK and a singly linked list LLL, you are supposed to reverse the links of every KKK elements on LLL. For example, given LLL being 1→2→3→4→5→6, if K=3K = 3K=3, then you must output 3→2→1→6→5→4; if K=4K = 4K=4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive NNN (≤105\le 10^5≤10​5​​) which is the total number of nodes, and a positive KKK (≤N\le N≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

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

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

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

Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 struct Node
 5 {
 6     int data;
 7     int adr,next;
 8 }t;
 9 int N,head,K;
10 map<int,Node> Input;//用map储存输入序列，地址作为键值，方便通过地址（head，next）找到相关点
11 stack<Node> Sta,T;
12 vector<Node> Result(N);//储存结果链表
13 vector<Node>::iterator it;
14 int main()
15 {
16     /*输入*/
17     scanf("%d %d %d",&head,&N,&K);
18     for(int i=0;i<N;i++)
19     {
20         scanf("%d %d %d",&t.adr,&t.data,&t.next);
21         Input[t.adr]=t;
22     }
23     /*处理*/
24     int num=0;
25     t=Input.find(head)->second;//找到第一个点
26     while(num++<N)//对n个元素入栈处理
27     {
28         int flag=t.next;
29         Sta.push(t);
30         if(t.next!=-1)
31             t=Input.find(t.next)->second;
32         if(num%K==0)//满K个元素，逆序出栈放入Result
33         {
34             while(!Sta.empty())
35             {
36                 Result.push_back(Sta.top());
37                 Sta.pop();
38             }
39             if(flag==-1)//最后一个结点刚好组成最后K个，结束
40                 break;
41             continue;
42         }
43         else if(flag==-1)//最后一个结点多余，将栈中的点出，入，出栈后恢复正序放入Result，结束
44         {
45             while(!Sta.empty())
46             {
47                 T.push(Sta.top());
48                 Sta.pop();
49             }
50             while(!T.empty())
51             {
52                 Result.push_back(T.top());
53                 T.pop();
54             }
55             break;
56         }
57     }
58     /*输出*/
59     it=Result.begin();
60     printf("%05d %d ",it->adr,it->data);
61     it++;
62     for(;it!=Result.end();it++)
63     {
64         printf("%05d\n%05d %d ",it->adr,it->adr,it->data);
65     }
66     printf("-1\n");
67     Input.clear();
68     Result.clear();
69     return 0;
70 }

 

 posted on 2017-04-17 16:24  theFresh  阅读(546)  评论(0)  编辑  收藏  举报