题目
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 ( ≤ 1 0 5 ) N (≤10^5) 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
思路分析：
还应该考虑输出中有不再链表的结点，所以用sum计数。可以调用algorithm中的reverse函数。
代码：
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100010;
int head, idx, e[maxn], ne[maxn], n, k, list[maxn];
int main(){
cin >> head >> n >> k;
for(int i = 0; i < n; i++){
cin >> idx;
cin >> e[idx] >> ne[idx];
}
int sum = 0;
while(head != -1){
list[sum++] = head;
head = ne[head];
}
for(int i = 0; i < sum / k; i += k)
reverse(list + i, list + i + k);
for(int i = 0; i < sum - 1; i++)
printf("%05d %d %05d\n", list[i], e[list[i]], list[i + 1]);
printf("%05d %d -1", list[sum - 1], e[list[sum - 1]]);
return 0;
}