PAT 1074. Reversing Linked List (25)

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

简单的模拟题,以空间换时间：

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const int NUM=;

struct Node
{
    int address;
    int data;
    int next;
};

Node nodes[NUM];
vector<Node> list;
int main()
{
    int fnAddress,N,K;
    scanf("%d %d %d",&fnAddress,&N,&K);
    int i;
    for(i=;i<N;++i)
    {
        Node node;
        scanf("%d %d %d",&node.address,&node.data,&node.next);
        nodes[node.address]=node;
    }
    int address=fnAddress;
    while(address!=-)//去噪
    {
        list.push_back(nodes[address]);
        address=nodes[address].next;
    }
    int size=list.size();
    int round=size/K;
    int start,end;
    for(i=;i<=round;++i)
    {
        start=(i-)*K;
        end=i*K;
        reverse(list.begin()+start,list.begin()+end);
    }
    for(i=;i<size-;++i)
    {
        printf("%.5d %d %.5d\n",list[i].address,list[i].data,list[i+].address);
    }
    printf("%.5d %d %d\n",list[i].address,list[i].data,-);
    return ;
}