A1133 Splitting A Linked List (25 point(s))

链表静态表示address[a]=a; data[a];nextaddress[a]

1. 原文

Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

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 (≤103). 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 in [−105,105], and Next is the position of the next node. It is guaranteed that the list is not empty.

output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218

Sample output:

33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1

2. 解析

链表各部分用map<int,int> 表示 address，data，nextaddress.

链表存在断链情况，for(int i=first;i!=-1;i=nextaddress[i]) 保证链

存入结果

for (int i = first; i != -1; i=nextAddress[i])
{
	if (data[i]<0)
	{
		v.push_back(i);
	}	
}

3. AC代码

#include<cstdio>
#include<map>
#include<vector>
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]);
	}
	for (int i = first; i != -1; i=nextAddress[i])
	{
		if (data[i]<0)
		{
			v.push_back(i);
		}	
	}
	for (int i = first; i != -1; i=nextAddress[i])
	{
		if (data[i]>=0&&data[i]<=k)
		{
			v.push_back(i);
		}	
	}
	for (int i = first; i != -1; i=nextAddress[i])
	{
		if (data[i]>k)
		{
			v.push_back(i);
		}	
	}
	for (int i = 0; i < v.size(); ++i)
	{
		if (i<v.size()-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;
}