PAT A1165 Block Reversing (25point(s))

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Given a singly linked list L. Let us consider every K nodes as a block (if there are less than K nodes at the end of the list, the rest of the nodes are still considered as a block). Your job is to reverse all the blocks in L. For example, given L as 1→2→3→4→5→6→7→8 and K as 3, your output must be 7→8→4→5→6→1→2→3.

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 size of a block. 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 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218

Sample Output:

71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1

解答

#include<iostream>
#include<unordered_map>
#include<vector>

using namespace std;

typedef struct {
	int add, data, next;
}node;

int main()
{
	int head, n, k, add, data, next;
	unordered_map<int, node> list;
	scanf("%d %d %d", &head, &n, &k);
	for (int i = 0; i < n; i++) {
		scanf("%d %d %d", &add, &data, &next);
		list[add] = node{ add, data, next };
	}
	int p = head;
	vector<node> li, ans;
	while (p != -1) {
		li.push_back(list[p]);
		p = list[p].next;
	}
	int re = li.size() % k, size = li.size();
	int from = size - 1;
	if (re != 0) {
		for (int j = 0; j < re; j++)
			ans.push_back(li[size - re + j]);
		from -= re;
	}
	from -= k;
	for (; from >= -1; from -= k) {
		for (int j = 1; j <= k; j++) 
			ans.push_back(li[from + j]);
	}
	for (int i = 0; i < size - 1; i++)
		printf("%05d %d %05d\n", ans[i].add, ans[i].data, ans[i + 1].add);
	printf("%05d %d -1", ans[size - 1].add, ans[size - 1].data);
	return 0;
}