A1074 Reversing Linked List (25 point(s))

反转链表

1. 原文

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

2. 解析

链表地址存入vector，地址反转reverse(v.begin()+i,v.begin()+i+k);

3. AC代码

#include<cstdio>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;
map<int,int> address,data,nextaddress;
vector<int> v;
int main()
{
	int first,n,k;
	scanf("%d%d%d",&first,&n,&k);
	int a;
	for (int i = 0; i < n; ++i)
	{
		scanf("%d",&a);
		address[a]=a;
		scanf("%d%d",&data[a],&nextaddress[a]);
	}
	int cnt=0;
	for (int i = first; i != -1; i=nextaddress[i])
	{
		cnt++;
		v.push_back(i);
	}
	for (int i = 0; i <cnt-cnt%k ; i+=k)
	{
		reverse(v.begin()+i,v.begin()+i+k);
	}
    for (int i = 0; i < cnt; ++i)
    {
    	if (i<cnt-1)
    	{
    		printf("%05d %d %05d\n",
               address[v[i]],data[v[i]],address[v[i+1]]);
    	}else{
    		printf("%05d %d -1\n",address[v[i]],data[v[i]]);
    	}
    }
	return 0;
}