PAT A1074 Reversing Linked List（25 分）

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

 代码：

#include
#include
using namespace std;

const int MAXV = 100010;

struct Node{
	int address;
	int data;
	int next;
	int order;
	Node(){
		order = 100011;
	}
}node[MAXV];

bool cmp(Node a, Node b){
	return a.order < b.order;
}

int main(){
	int address, n, k;
	scanf("%d %d %d", &address, &n, &k);
	for(int i = 0; i < n; i++){
		int adr, data, next;
		scanf("%d %d %d", &adr, &data, &next);
		node[adr].address = adr;
		node[adr].data = data;
		node[adr].next = next;
	}
	int tempaddress = address;
	int count = 0;				//有效节点个数
	for(int j = 0; j < n; j++){
		node[tempaddress].order = count++;
		tempaddress = node[tempaddress].next;
		if(tempaddress == -1) break;
	}
	sort(node, node + MAXV, cmp);
	int total_reverse = count;	//
	int time_reverse = 0;	//第几个需要反转的链表
	while(total_reverse / k >0){
		reverse(node + time_reverse * k, node + (time_reverse + 1) * k);
		time_reverse++;
		total_reverse -= k;
	}
	for(int t = 0; t < count; t++){
		if(t == count-1) printf("%05d %d -1\n", node[t].address, node[t].data);
		else printf("%05d %d %05d\n", node[t].address, node[t].data, node[t+1].address);
	}
	return 0;
}