02-1. Reversing Linked List (25)

http://www.patest.cn/contests/mooc-ds/02-1

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 
using namespace std;
#define MAXN 100001
typedef struct{
	int addr;
	int data;
	int next;
}Node;

Node nodes[MAXN];
vector list;
int main(){
	int firstAdd, n, k;
	scanf("%d%d%d", &firstAdd, &n, &k);
	while(n--){
		Node nn;
		scanf("%d%d%d", &nn.addr, &nn.data, &nn.next);
		nodes[nn.addr] = nn;
	}
	int address = firstAdd;
	while(address != -1){
		list.push_back(nodes[address]);
		address = nodes[address].next;
	}
	int length = list.size();
	int round = length/k;
	for(int i = 1; i <= round; ++i){
		int start = (i-1)*k;
		int end = i*k;
		reverse(list.begin() + start, list.begin() + end);
	}
	for(int i = 0; i < length-1; ++i){
		printf("%05d %d %05d\n", list[i].addr, list[i].data, list[i+1].addr);
	}
	printf("%05d %d %d\n",list[length-1].addr, list[length-1].data, -1);
	return 0;
}