A1074 Reversing Linked List [链表高级倒置]

分块先分块输出再讨论输出每块最后一个点，看这个点是否在最后一块内。

#include<vector>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<set>
#include<map>
#include<cstring>
#include<string>
#include<queue>
#include<array>
#include<stack>
using namespace std;
const int maxn = 100010;
struct Node
{
	int address,data,next;
	bool flag;
	int order;
}node[maxn];
bool cmp(Node a, Node b)
{
	return a.order < b.order;
}
int main()
{
	for (int i = 0; i < maxn; i++)
	{
		node[i].order = maxn;
	}
	int begin, n, k;
	cin >> begin >> n >> k;
	int address, data, next;
	for (int i = 0; i < n; i++)
	{
		cin >> address >> data >> next;
		node[address].address = address;
		node[address].next = next;
		node[address].data = data;
	}
	int p = begin, count = 0;
	while (p != -1)
	{
		node[p].order = count++;
		p = node[p].next;
	}
	sort(node, node + maxn, cmp);
	n = count;
	for (int i = 0; i < n / k; i++)
	{
		for (int j = (i + 1) * k - 1; j > i* k; j--)
		{
			printf("%05d %d %05d
", node[j].address, node[j].data, node[j - 1].address);
		}
		//讨论每一块最后一个点
		printf("%05d %d ", node[i * k].address, node[i * k].data);
		if (i < n / k - 1)
		{
			printf("%05d
", node[(i + 2) * k - 1].address);
		}
		else
		{
			if (n % k == 0)
				cout << "-1" << endl;
			else
			{
				printf("%5d
", node[(i + 1) * k].address);
				for (int i = n / k * k; i < n; i++)
				{
					printf("%05d %d ", node[i].address, node[i].data);
					if (i < n - 1)
					{
						printf("%05d
", node[i].next);
					}
					else
					{
						printf("-1
");
					}
				}
			}
		}
	}
}

#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1000001;
struct Node
{
	int address, data, next;
	int order;
}node[maxn];
bool cmp(Node a, Node b)
{
	return a.order < b.order;
}
int main()
{
	for (int i = 0; i < maxn; i++)
	{
		node[i].order = maxn;
	}
	int begin, n, k;
	cin >> begin >> n >> k;
	int address, data, next;
	for (int i = 0; i < n; i++)
	{
		cin >> address >> data >> next;
		node[address].address = address;
		node[address].data = data;
		node[address].next = next;
	}
	int p = begin, count = 0;
	while (p != -1)
	{
		node[p].order = count++;
		p = node[p].next;
	}
	n = count;//顺应题意
	sort(node, node + maxn, cmp);
	for (int i = 0; i <n/k; i++)
	{
		for (int j = (i + 1) * k - 1; j > i* k; j--)
		{
			printf("%05d %d %05d
", node[j].address, node[j].data, node[j - 1].address);
		}
		printf("%05d %d ", node[i * k].address, node[i * k].data);
		if (i < n/k - 1)
		{
			printf("%05d
", node[(i + 2) * k - 1].address);
		}
		else
		{
			if (n % k == 0)
				cout << "-1" << endl;
			else
			{
				printf("%05d
", node[(i + 1) * k].address);
				for (int q = (i + 1) * k; q < n; q++)
				{
					printf("%05d %d ", node[q].address, node[q].data);
					if (q < n - 1)
					{
						printf("%05d
", node[q].next);
					}
					else
					{
						printf("-1
");
					}
				}
			}
		}
	}

}