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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1

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#include<iostream>#include<cstdio>using namespace std;struct List{int head;int num;int next;};struct List L[100000];struct List L2[100000];int start,n,k;int a,b,c;int index1=0;void sswap(int a,int b){     struct List t;    for (int i=a,j=b;i<=j;i++,j--)    {         t=L2[i];L2[i]=L2[j];L2[j]=t;    }}int main(){    cin>>start>>n>>k;    int s=start;    for (int i=0;i<n;i++)    {        cin>>a>>b>>c;        L[a].head=a;        L[a].num=b;        L[a].next=c;    }    while(1)    {       L2[index1++]=L[s];       if(L[s].next==-1)        break;       s=L[s].next;    }    for(int i=0;i+k-1<index1;i+=k)        sswap(i,i+k-1);    for (int i=0;i<index1-1;i++)        printf("%05d %d %05d\n",L2[i].head,L2[i].num,L2[i+1].head);     printf("%05d %d -1\n",L2[index1-1].head,L2[index1-1].num);    return 0;}