1074 Reversing Linked List (25 point(s))

1074 Reversing Linked List (25 point(s))

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 (≤10​5​​) 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

链表。

思路很简单，主要考虑两点：

1. 可能存在不在链表中的无效结点；

2. 注意输出格式，比如补零。

一定要读题仔细，看着样例分析的过程中，以为要先排序（样例正好是这样的），然后再每K个元素翻转，结果浪费了整整半个小时。 

#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int MAX = 1e5+7;
int data[MAX];
int nextAdd[MAX];
int main(void){
	int root,N,K;cin>>root>>N>>K;
	int address,key,next;
	for(int i=0;i<N;i++){
		cin>>address>>key>>next;
		data[address]=key;
		nextAdd[address]=next;
	}
	vector<int> v;
	while(root!=-1){
		v.push_back(root);
		root = nextAdd[root];
	}
	for(int i=0;i+K<=v.size();i=i+K){
		reverse(v.begin()+i,v.begin()+i+K);
	}
	for(int i=0;i<v.size();i++){
		if(i==v.size()-1) printf("%05d %d -1\n",v[i],data[v[i]]);
		else printf("%05d %d %05d\n",v[i],data[v[i]],v[i+1]);
	}
	return 0;
}