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

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100100;
struct Node{
    int address,data,next;
    int order;
}node[maxn];

bool cmp(Node a,Node b){
    return a.order < b.order;
}
int main(){
    int i,address;
    for(i = 0; i < maxn; i++){
        node[i].order = maxn;
    }
    int begin,n,k;  //起始节点地址，节点数目，分组数 
    scanf("%d%d%d",&begin,&n,&k);
    
    for(i = 0; i < n; i++){
        scanf("%d",&address);
        scanf("%d%d",&node[address].data,&node[address].next);
        node[address].address = address;
    }
    int p = begin,count = 0;
    while(p != -1){
        node[p].order = count++;
        p = node[p].next;
    }
    sort(node,node+maxn,cmp);
    n = count; //因为count=0占一个有效节点，退出循环时，count值就是有效节点 
    for(i = 0; i < n/k; i++){ //枚举完整的n/k块 
       for(int j = (i+1)*k - 1; j > i*k; j-- ){ //每块的第i个倒着输出，剩余最后一个节点 
           printf("%05d %d %05d\n",node[j].address,node[j].data,node[j-1].address);
       }
       printf("%05d %d ",node[i*k].address,node[i*k].data); //每块的最后一个节点的前两项数据 
        if(i < n/k -1)  { //如果是非最后一块节点 
            printf("%05d\n",node[(i+2)*k-1].address);
        }else{   //如果是最后一块节点 
            if(n % k == 0) printf("-1\n");  //刚好除整 
            else{   //如果最后一个节点不规则 
            printf("%05d\n",node[(i+1)*k].address);
            for(i = n/k*k; i < n; i++){
                printf("%05d %d ",node[i].address,node[i].data);
                if(i < n - 1){
                    printf("%05d\n",node[i+1].address);
                }else{
                    printf("-1\n");
                 }// else
            } // for(i)
           }//else
        }//else
    } //for(i)
   
    return 0;
}