PAT A1074 Reversing Linked List （25 分）——链表，vector，stl里的reverse

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 (≤10​5​​) 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 <stdio.h>
 #include <map>
 #include <algorithm>
 #include <iostream>
 #include <string>
 #include <math.h>
 #include <vector>
 using namespace std;
 const int maxn = ;
 struct node{
     int addr;
     int data;
     int next;
 }nodes[maxn];
 vector<node> v;
 int main(){
     int st, n, k, count = ;
     cin >> st >> n >> k;
     for (int i = ; i < n; i++){
         int start, data, next;
         cin >> start >> data >> next;
         nodes[start].addr = start;
         nodes[start].data = data;
         nodes[start].next = next;
     }
     while (st != -){
         v.push_back(nodes[st]);
         st = nodes[st].next;
         count++;
     }
     for (int i = ; i+k <= count; i = i + k){
         reverse(v.begin() + i, v.begin() + i + k);
     }
     for (int i = ; i < count-; i++){
         v[i].next = v[i + ].addr;
     }
     v[count-].next = -;
     for (int i = ;i < count-; i++){
         printf("%05d %d %05d\n", v[i].addr, v[i].data, v[i].next);

     }
     printf("%05d %d %d\n", v[count - ].addr, v[count - ].data, v[count - ].next);
     system("pause");
 }

注意点：同B1025，链表题，都会要你重新排序输出，注意可以用vector实现链表，不要用静态数组，保存节点时一定要保存自己的地址。反转可以直接使用algorithm里的reverse