PAT1074

PAT1074 Reversing Linked List (25 分)

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 (≤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 
using namespace std;
struct Node{
	int ad;
	int data;
	int next;
} node[100001];
int main() {
	vector<Node> v;
	int s, n, k, i;
	int sq;
	scanf("%d %d %d", &s, &n, &k);
	for (i = 0; i < n;++i)
	{
		scanf("%d ", &sq);
		node[sq].ad = sq;
		scanf("%d %d", &node[sq].data,&node[sq].next);
	}
	sq = 0;
	for (i = s;i!=-1; i =node[i].next)
		{
			v.push_back(node[i]);
			sq++;
		}
	for (auto j=v.begin();j+k<=v.end();j+=k)
		reverse(j,j+k);
	for (i = 0; i < v.size()-1; ++i)
	{
		printf("%05d %d %05d\n", v[i].ad,v[i].data,v[i+1].ad);
	}
	printf("%05d %d -1\n", v[i].ad, v[i].data);
	return 0;
}