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 4 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218Sample Output:
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1
1 #include <cstdio> 2 #include <cstring> 3 #include <string> 4 #include <queue> 5 #include <stack> 6 #include <cmath> 7 #include <iostream> 8 #include <algorithm> 9 using namespace std; 10 struct node{ 11 int now,next,data; 12 }; 13 node mem[100005]; 14 vector<node> 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<num;i++){ 21 scanf("%d %d %d",&now,&data,&next); 22 mem[now].now=now; 23 mem[now].data=data; 24 mem[now].next=next; 25 } 26 now=h; 27 while(now!=-1){ 28 v.push_back(mem[now]); 29 now=mem[now].next; 30 } 31 int length=v.size(); 32 int round=length/k; 33 for(i=0;i<round;i++){ 34 reverse(v.begin()+i*k,v.begin()+i*k+k); 35 } 36 for(i=0;i<length-1;i++){ 37 printf("%05d %d %05d ",v[i].now,v[i].data,v[i+1].now); 38 } 39 printf("%05d %d %d ",v[i].now,v[i].data,-1);//注意全反的情况 40 return 0; 41 }