1074. Reversing Linked List (25)

模拟题，注意当k == 1 与 k == n时情况

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <map>

using namespace std;

const int N = 100005;

struct Node
{
	int pre;
	int value;
	int lat;
}node[N];

int order[N];
int size;
map<int, int> pre2idx;

void solve1(int fir, int n)
{
	map<int, int>::iterator it = pre2idx.find(fir);
	order[size++] = it->second;

	while (node[it->second].lat != -1)
	{
		it = pre2idx.find(node[it->second].lat);
		order[size++] = it->second;
	}
}

void solve2(int k, int n)
{
	int bound = n / k * k;

	int idx = k;
	while (idx <= bound)
	{
		int boundt = idx;
		int boundb = idx - k + 1;
		while (boundt >= boundb)
		{
			int next = 0;
			if (boundt > boundb) next = boundt - 1;
			else
			{
				if (idx < bound)
					next = idx + k;
				else if (idx == bound)
				{
					if (bound < n)
						next = idx + 1;
					else if (bound == n)
					{
						printf("%05d %d -1\n", node[order[boundt]].pre, node[order[boundt]].value);
						break;
					}
				}
			}
			printf("%05d %d %05d\n", node[order[boundt]].pre, node[order[boundt]].value, node[order[next]].pre);
			boundt--;
		}
		idx += k;
	}
	bound++;
	while (bound <= n)
	{
		if (node[order[bound]].lat != -1)
			printf("%05d %d %05d\n", node[order[bound]].pre, node[order[bound]].value, node[order[bound]].lat);
		else 	printf("%05d %d -1\n", node[order[bound]].pre, node[order[bound]].value);
		bound++;
	}

}

int main()
{
	int fir, n, k;

	while (scanf("%d%d%d", &fir, &n, &k) != EOF)
	{
		pre2idx.clear(); size = 1;
		for (int i = 0; i < n; i++)
		{
			scanf("%d%d%d", &node[i].pre, &node[i].value, &node[i].lat);
			pre2idx.insert(make_pair(node[i].pre, i));
		}

		solve1(fir, n);

		solve2(k , size - 1);

	}
	return 0;
}

/*
00100 6 6
00000 4 99999
00100 1 -1
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
*/