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 len >h 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 resu< ing 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

• 题意：给定一个静态链表，首地址begin1，结点数n，排序长度k。对于每一个节点，给定该节点的地址、数据以及该节点下一个节点的地址。
• 定义一个结构体，其中包括节点的地址、数据、下一节点的地址以及当前节点在链表中的序号。
• 首先按照节点id从小到大排序，所以初始化的时候需要让节点id均初始化为maxn，即让一些没有出现的节点放到筛掉。
• 按照要求进行排序即可。

``````#includecstdio
#includealgorithm
using namespace std;
const int maxn=100001;
struct Node{
int data;
int next;
int id;	//分别表示节点地址，数据，下一节点的地址，节点的id
}node[maxn];
int begin1,n,k;
bool cmp1(Node a,Node b)
{
return a.idb.id;

}
bool cmp2(Node a,Node b)
{
return a.idb.id;
}
void solve()
{
for(int i=0;imaxn;i++)
node[i].id=maxn;
scanf("%d %d %d",begin1,n,k);
for(int i=0;in;i++)
{
int start;
scanf("%d",start);
//node[start].id=i;
scanf("%d %d",node[start].data,node[start].next);
}
int index=0;
while(begin1!=-1)
{
node[begin1].id=index++;
begin1=node[begin1].next;
}
sort(node,node+maxn,cmp1);
/*需要去除无效节点*/
for(int i=0;iindex/k;i++)	//改成n会有一个测试样例通不过
{
sort(node+i*k,node+(i+1)*k,cmp2);
}
for(int i=0;iindex;i++)
{
if(i!=index-1)
{
}
else
{
}
}
}
int main()
{
solve();
return 0;
}
``````
• 1:51:00