Pat Grade 1074. Reversing Linked List (25)

Given a constant k and a singly linked list L, you is 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 the one test case. The first line contains the address of the first node, a positive N (<=) which are the total number O F nodes, and a positive K (<=n) which are 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
The where Address is the position of the node, Data are 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 was 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 <cstdio> using namespace std;
#include <vector> const int max=100000+10;

int Node[max],next[max],previous[max];
    int main () {int head,n,k;

    scanf ("%d%d%d", &head,&n,&k);
    int address,data,next_address;
        for (int i=0;i<n;i++) {scanf ("%d%d%d", &address,&data,&next_address);
            if (address!=-1) {node[address]=data;
        next[address]=next_address;
        } if (Address==head) previous[address]=-1;
    if (next_address!=-1) previous[next_address]=address;
    } vector<int> Index;
    int cnt=0;//The number of nodes in the list for (int i=head;i!=-1;i=next[i]) {index.push_back (i);
        printf ("previous=%d\n", Previous[i]);
    cnt++;
    } int round=cnt/k;
    int m=k-1;
    int last=-1;
        while (m<cnt) {int i=m;
        for (i;i>m-k+1;i--) {printf ("%05d%d%05d\n", Index[i],node[index[i]],previous[index[i]]); } if (m==cnt-1) printf ("%05d%d%d\n ", index[i],node[index[i]],-1);
        else if (m+k>=cnt) printf ("%05d%d%05d\n", index[i],node[index[i]],index[m+1]);
        else printf ("%05d%d%05d\n", index[i],node[index[i]],index[m+k]);
    M+=k; 
        } for (int i=m-k+1;i<cnt;i++) {if (i!=cnt-1) printf ("%05d%d%05d\n", Index[i],node[index[i]],next[index[i]]);
    else printf ("%05d%d%d\n", index[i],node[index[i]],-1);
} return 0; }