1074 Reversing Linked List

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

和乙级的1025一样：1025 反转链表 (25 分)_Brosto_Cloud的博客-CSDN博客 

Data 和 Next 数组的下标 i 都是地址，表示对应地址的数据和链接的下一个地址，list数组下标 i 是顺序，代表节点地址，每个地址对应的数据都是不变的，改变的只是链接的下一个地址，所以对list 数组每k个进行反转后直接输出 list[i]和list[i+1]就是对应的地址顺序，data[i]是对应的数据。

#include 
#include 
#include 
using namespace std;
int Data[100010], list[100010], Next[100010];
int n, k, addr, t, sum;
 
int main() {
	cin >> addr >> n >> k;
	for (int i = 0; i < n; i++) {
		cin >> t;
		cin >> Data[t] >> Next[t];
	}
	while (addr != -1) {
		list[sum++] = addr;
		addr = Next[addr];
	}
	for (int i = 0; i < (sum - sum % k); i += k) {
		reverse(list + i, list + i + k);
	}
	for (int i = 0; i < sum - 1; i++) {
		cout << setfill('0') << setw(5) << list[i] << ' ' << Data[list[i]] << ' ' << setfill('0') << setw(
		         5) << list[i + 1] << endl;
	}
	cout << setfill('0') << setw(5) << list[sum - 1] << ' ' << Data[list[sum - 1]] << ' ' << -1;
	return 0;
}