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
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 using namespace std; 5 typedef struct NODE{ 6 int lt, rt; 7 int valid; 8 int data, rk; 9 }node; 10 node nds[100001]; 11 bool cmp(node a, node b){ 12 if(a.valid != b.valid) 13 return a.valid > b.valid; 14 else return a.rk < b.rk; 15 } 16 bool cmp2(node a, node b){ 17 return a.rk > b.rk; 18 } 19 int main(){ 20 int N, K, head; 21 scanf("%d%d%d", &head, &N, &K); 22 int temp; 23 for(int i = 0; i < N; i++){ 24 scanf("%d", &temp); 25 nds[temp].lt = temp; 26 scanf("%d%d", &nds[temp].data, &nds[temp].rt); 27 } 28 int pt = head, cnt = 0; 29 while(pt != -1){ 30 nds[pt].valid = 1; 31 nds[pt].rk = cnt; 32 cnt++; 33 pt = nds[pt].rt; 34 } 35 sort(nds, nds + 100001, cmp); 36 for(int i = 0; i + K <= cnt; i += K){ 37 sort(nds + i, nds + i + K, cmp2); 38 } 39 if(cnt == 0) 40 printf("-1"); 41 for(int i = 0; i < cnt; i++){ 42 if(i < cnt - 1) 43 printf("%05d %d %05d ", nds[i].lt, nds[i].data, nds[i + 1].lt); 44 else printf("%05d %d -1 ", nds[i].lt, nds[i].data); 45 } 46 cin >> N; 47 return 0; 48 }
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