PAT (Advanced Level) Practise 1074 Reversing Linked List (25)

1074. Reversing Linked List (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (<= 105) 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

先把地址存到一个数组里，然后直接反转，在按顺序输出。
#include<cstdio>
#include<stack>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
const int INF = 0x7FFFFFFF;
const int maxn = 1e5 + 15;
int root, nt[maxn], v[maxn], n, k, x;
int f[maxn], tot;

int main()
{
scanf("%d%d%d", &root, &n, &k);
while (n--)
{
scanf("%d", &x);
scanf("%d", &v[x]);
scanf("%d", &nt[x]);
}
for (int i = root; i != -1; i = nt[i])
{
f[tot++] = i;
}
for (int i = 0; i + k <= tot; i += k)
{
for (int j = 0; j < k; j++)
{
if (i + j>i + k - j - 1) break;
swap(f[i + j], f[i + k - j - 1]);
}
}
for (int i = 0; i < tot; i++)
{
if (i < tot - 1) printf("%05d %d %05d\n", f[i], v[f[i]], f[i + 1]);
else printf("%05d %d -1\n", f[i], v[f[i]]);
}
return 0;
}