1074 Reversing Linked List （25 分）(链表)

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 (≤10

5

) 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

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <string.h>
#include <cstdio>
using namespace std;
struct Node{
    int data,next;
}node[100000];
int main(){
    int head,n,k,ad;
    cin>>head>>n>>k;
    for(int i=0;i<n;i++){
        scanf("%d",&ad);
        scanf("%d%d",&node[ad].data,&node[ad].next);
    }
    int now=head,a[100000],sum=0;
    while(now!=-1){
        a[sum++]=now;
        now=node[now].next;
    }
    for(int i=0;i<=sum-k;i+=k)
        reverse(a+i,a+i+k);
    for(int i=0;i<sum-1;i++)
        printf("%05d %d %05d\n",a[i],node[a[i]].data,a[i+1]);
    printf("%05d %d -1\n",a[sum-1],node[a[sum-1]].data);
    return 0;
}