A1074. 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

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 typedef struct NODE{
     int lt, rt;
     int valid;
     int data, rk;
 }node;
 node nds[];
 bool cmp(node a, node b){
     if(a.valid != b.valid)
         return a.valid > b.valid;
     else return a.rk < b.rk;
 }
 bool cmp2(node a, node b){
     return a.rk > b.rk;
 }
 int main(){
     int N, K, head;
     scanf("%d%d%d", &head, &N, &K);
     int temp;
     for(int i = ; i < N; i++){
         scanf("%d", &temp);
         nds[temp].lt = temp;
         scanf("%d%d", &nds[temp].data, &nds[temp].rt);
     }
     int pt = head, cnt = ;
     while(pt != -){
         nds[pt].valid = ;
         nds[pt].rk = cnt;
         cnt++;
         pt = nds[pt].rt;
     }
     sort(nds, nds + , cmp);
     for(int i = ; i + K <= cnt; i += K){
         sort(nds + i, nds + i + K, cmp2);
     }
     if(cnt == )
         printf("-1");
     for(int i = ; i < cnt; i++){
         if(i < cnt - )
             printf("%05d %d %05d\n", nds[i].lt, nds[i].data, nds[i + ].lt);
         else printf("%05d %d -1\n", nds[i].lt, nds[i].data);
     }
     cin >> N;
     return ;
 }

总结：

1、注意链表节点的初始顺序不是读入的顺序，而是在读完之后，从给定的首地址遍历一遍的顺序。

2、静态链表题一般都会在输入中放入无效节点，需要过滤掉。注意全空的情况。

3、本题可以在第一次遍历合法节点的时候，给每个节点都编号0、1、2......，然后按照valid和节点编号排序，这样可以将所有节点聚集起来且按照链表本身的顺序依次存放。之后第二遍遍历，每K个从大到小再排一次序即可得到最终的顺序。

4、每个节点自己的地址是始终不变的。

5、只有到了最后一步才可以把所有合法节点聚集到数组前部。

6、测试点

　　00000 6 3
　　00000 1 11111
　　11111 2 22222
　　22222 3 -1
　　33333 4 44444
　　44444 5 55555
　　55555 6 -1