02-线性结构1. 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 33218Sample Output:
00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 9 using namespace std;10 struct node{11 int now,next,data;12 };13 node mem[100005];14 vector v;15 int main(){16 //freopen("D:\\INPUT.txt","r",stdin);17 int h,now,data,next,num,k,temp;18 scanf("%d %d %d",&h,&num,&k);19 int i;20 for(i=0;i