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

 #include<bits/stdc++.h>
 using namespace std;

 struct Node
 {
     int data;
     int adr,next;
 }t;
 int N,head,K;
 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
 stack<Node> Sta,T;
 vector<Node> Result(N);//储存结果链表
 vector<Node>::iterator it;
 int main()
 {
     /*输入*/
     scanf("%d %d %d",&head,&N,&K);
     for(int i=;i<N;i++)
     {
         scanf("%d %d %d",&t.adr,&t.data,&t.next);
         Input[t.adr]=t;
     }
     /*处理*/
     int num=;
     t=Input.find(head)->second;//找到第一个点
     while(num++<N)//对n个元素入栈处理
     {
         int flag=t.next;
         Sta.push(t);
         if(t.next!=-)
             t=Input.find(t.next)->second;
         if(num%K==)//满K个元素，逆序出栈放入Result
         {
             while(!Sta.empty())
             {
                 Result.push_back(Sta.top());
                 Sta.pop();
             }
             if(flag==-)//最后一个结点刚好组成最后K个，结束
                 break;
             continue;
         }
         else if(flag==-)//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
         {
             while(!Sta.empty())
             {
                 T.push(Sta.top());
                 Sta.pop();
             }
             while(!T.empty())
             {
                 Result.push_back(T.top());
                 T.pop();
             }
             break;
         }
     }
     /*输出*/
     it=Result.begin();
     printf("%05d %d ",it->adr,it->data);
     it++;
     for(;it!=Result.end();it++)
     {
         printf("%05d\n%05d %d ",it->adr,it->adr,it->data);
     }
     printf("-1\n");
     Input.clear();
     Result.clear();
     return ;
 }