1074. Reversing Linked List (25)

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本文介绍了一个算法问题：如何按每K个元素反转单链表L，并提供了完整的C++实现代码。输入包括链表的第一个节点地址、节点总数及反转长度K。文章详细展示了处理流程与输出格式。

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1074. Reversing Linked List (25)
时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 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 N (<= 105) which is the total number of nodes, and a positive K (<=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 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, 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

#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include <functional>
#include <iostream>
#include <queue>

using namespace std;
const int MaxN = 100010;

struct Slinklist
{
    int data;
    int nextAddr;
}SLT[MaxN] = {-1};

struct info
{
    int Addr;
    int data;
    int nextAddr;
}ans[MaxN];

int AddrStore[MaxN], n = 0;

int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif

    //std::ios::sync_with_stdio(false);

    int startAddr, N, K, Addr;
    scanf("%d %d %d", &startAddr, &N, &K);

    while (N--)
    {
        scanf("%d", &Addr);
        scanf("%d %d", &SLT[Addr].data, &SLT[Addr].nextAddr);
    }

    Addr = startAddr;

    while (Addr != -1)
    {
        int idx = 0;
        for (int i = 0; i < K && Addr != -1; ++i)
        {
            AddrStore[i] = Addr;
            Addr = SLT[Addr].nextAddr;
            ++idx;
        }


        for (int i = 0; i < idx; ++i, ++n)
        {
            if (idx == K)
            {
                ans[n].data = SLT[AddrStore[idx - i - 1]].data;
                ans[n].Addr = AddrStore[idx - i - 1];
            }
            else
            {
                ans[n].data = SLT[AddrStore[i]].data;
                ans[n].Addr = AddrStore[i];
            }

            if (i != idx - 1)
            {
                if (K == idx)
                    ans[n].nextAddr = AddrStore[idx - i - 2];
                else
                    ans[n].nextAddr = SLT[AddrStore[i]].nextAddr;
            }
            else if (Addr != -1)
            {
                int tmp = Addr, i = 0;
                for (; i < K-1 && SLT[tmp].nextAddr != -1; ++i)
                {
                    tmp = SLT[tmp].nextAddr;
                }
                if (i == K - 1)
                    ans[n].nextAddr = tmp;
                else
                    ans[n].nextAddr = Addr;
            }
            else
                ans[n].nextAddr = -1;
        }
    }

    for (int i = 0; i < n; ++i)
    {
        printf("%05d %d ",ans[i].Addr,ans[i].data);
        if (ans[i].nextAddr != -1)
            printf("%05d\n", ans[i].nextAddr);
        else
            printf("-1\n");
    }

    return 0;
}