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.
给出一个常数K和单链表L，你需要把链表L上每K个元素进行反转，比如，给出L：1->2->3->4->5->6.如果K=3，那么你需要输出3->2->1->6->5->4.如果K=4，你需要输出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.
每一个测试样例，第一行包含了第一个结点的地址，一个正数N(≤10^5)代表结点总量，一个正数K(≤N)代表需要反转的子链表长度。结点的地址是一个5位非负整数，NULL用-1来代表
之后有N行，每一行描述了一个结点，格式如图
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.
对于每一个测试样例，输出结果链表，每一个结点占一行，输出格式和输入格式相同
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100010;
struct Node{
int address,data,next;
int order;//结点在链表上的序号，无效节点记为maxn
}node[maxn];
bool cmp(Node a,Node b){
return a.order<b.order;
}
int main(){
for(int i=0;i<maxn;i++){
node[i].order=maxn;
}
int begin,n,K,address;
scanf("%d%d%d",&begin,&n,&K);
for(int 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;
for(int i=0;i<n/K;i++){
for(int j=(i+1)*K-1;j>i*K;j--){
printf("%05d %d %05d\n",node[j].address,node[j].data,node[j-1].address);
}
//下面是每一块最后一个结点的next地址处理
printf("%05d %d ",node[i*K].address,node[i*K].data);
if(i<n/K-1){//如果不是最后一块，就指向下一块的最后一个结点
printf("%05d\n",node[(i+2)*K-1].address);
}else{
if(n%K==0){//恰好是最后一个结点，输出-1
printf("-1\n");
}else{//剩下的不完整的块按原先顺序输出
printf("%05d\n",node[(i+1)*K].address);
for(int 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");
}
}
}
}
return 0;
}