PAT-B 1025. 反转链表(同PAT 1074. Reversing Link)

同PAT 1074. Reversing Link

按每单元 k 个结点来反转链表。

如，链表为1 -> 2 -> 3 -> 4 -> 5 -> 6，并且k = 4时

结果应为4 -> 3 -> 2 -> 1 -> 5 -> 6

1. 注意到，最后除不尽的那个单元是不翻转的。一开始题目没认真读，囧。

2. 测试点6考察的是输入的结点有的不在链表上的结果。我用dfs做时，就需要重新计算结点总数来限制搜索的深度。

代码：

#include <cstdio>
#include <vector>
#include <iostream>
#include <cmath>

using namespace std;

struct Node
{
	Node() {}
	Node(int data, int next): m_data(data), m_next(next) {}
	int m_data;
	int m_next;
};

Node node[100010];
int head, n, k;
int add, data, nexx, now;

int dfs(const vector<int>& vt, int level)
{
	if (level == 0)
	{
		head = vt.back();
	}

	vector<int> new_vt;
	if (level < n/k -1)
	{
		int now = node[vt.back()].m_next;
		for (int i = 0; i < k; ++ i)
		{
			new_vt.push_back( now );
			now = node[now].m_next;
		}
	}

	int end = node[vt[k-1]].m_next;
	for (int i = k-1; i > 0; -- i)
	{
		node[vt[i]].m_next = vt[i-1];
	}

	if (level < n/k - 1)
	{
		node[vt[0]].m_next = dfs(new_vt, level+1);
	} else
	{
		node[vt[0]].m_next = end; // 此时end == 后面的链表头 或 -1
	}

	return vt.back();
}

int count_n()
{
	int cnt = 0;

	now = head;
	while (now != -1)
	{
		++ cnt;
		now = node[now].m_next;
	}

	return cnt;
}

int main()
{
	scanf("%d%d%d", &head, &n, &k);

	for (int i = 0; i < n; ++ i)
	{
		scanf("%d%d%d", &add, &data, &nexx);
		node[add].m_data = data;
		node[add].m_next = nexx;
	}

	n = count_n();

	if (n == 0)
	{
		return 0;
	}

	vector<int> vt;
	now = head;
	for (int i = 0; i < k; ++ i)
	{
		vt.push_back(now);
		now = node[now].m_next;
	}
	dfs(vt, 0);

	now = head;
	while (node[now].m_next != -1)
	{
		printf("%05d %d %05d\n", now, node[now].m_data, node[now].m_next);
		now = node[now].m_next;
	}
	printf("%05d %d %d\n", now, node[now].m_data, node[now].m_next);

	return 0;
}