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 (≤105le 10^5≤10​5​​) which is the total number of nodes, and a positive KKK (≤Nle 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=0;i<N;i++)
    {
        scanf("%d %d %d",&t.adr,&t.data,&t.next);
        Input[t.adr]=t;
    }
    /*处理*/
    int num=0;
    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!=-1)
            t=Input.find(t.next)->second;
        if(num%K==0)//满K个元素，逆序出栈放入Result
        {
            while(!Sta.empty())
            {
                Result.push_back(Sta.top());
                Sta.pop();
            }
            if(flag==-1)//最后一个结点刚好组成最后K个，结束
                break;
            continue;
        }
        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
        {
            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
%05d %d ",it->adr,it->adr,it->data);
    }
    printf("-1
");
    Input.clear();
    Result.clear();
    return 0;
}