PAT(甲级)1074

1074. Reversing Linked List (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 
#include 
#include 
#include 
#define SIZE 100005

using namespace std;

stack s,s1;
deque  q;

struct Iterm{
	int value;
	int Next;
};

Iterm input[SIZE];

int  main()
{
	int start, num,k;
	scanf("%d%d%d",&start,&num,&k);
	int addr,nextaddr,value;
	for(int i =0;i0;count--){
			int addr= s.top();
			s.pop();
			int value = s.top();
			s.pop();
			if(fix){
				printf("%05d\n",addr);
			}
			printf("%05d %d ",addr,value);
			fix =true;
			}
		}
	}
	while(!s.empty()){
		int addr=s.top();
		s.pop();
		s1.push(s.top());
		s.pop();
		s1.push(addr);
	}
	while(!s1.empty()){
	    if(fix)
	        printf("%05d\n",s1.top());
		int addr = s1.top();
		s1.pop();
		int value = s1.top();
		s1.pop();
		printf("%05d %d ",addr,value);
	}
	cout <<-1 <