【PAT】1052. Linked List Sorting - map/qsort用法

原题链接 http://pat.zju.edu.cn/contests/pat-a-practise/1052

1052. Linked List Sorting (25)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

A linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and a Next pointer to the next structure. Now given a linked list, you are supposed to sort the structures according to their key values in increasing order.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive N (< 10⁵) and an address of the head node, where N is the total number of nodes in memory and the address of a node is a 5-digit positive integer. NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Key Next

where Address is the address of the node in memory, Key is an integer in [-10⁵, 10⁵], and Next is the address of the next node. It is guaranteed that all the keys are distinct and there is no cycle in the linked list starting from the head node.

Output Specification:

For each test case, the output format is the same as that of the input, where N is the total number of nodes in the list and all the nodes must be sorted order.

Sample Input:

5 00001
11111 100 -1
00001 0 22222
33333 100000 11111
12345 -1 33333
22222 1000 12345

Sample Output:

5 12345
12345 -1 00001
00001 0 11111
11111 100 22222
22222 1000 33333
33333 100000 -1

题意： 从给出的节点中，找出一条链。然后将其按照其中的key从小到大排序显示。

代码如下：

#include <iostream>
#include <fstream>
#include <map>
#include <vector>
#include <iomanip>
using namespace std;

ifstream fin("in.txt");
//#define cin fin

struct Node
{
	int key;
	int next;
	Node(){}
	Node(int addr,int k){key = k;next = addr;}
};

int cmp(const void* A,const void* B)
{
	Node* aa = (Node*)A;
	Node* bb = (Node*)B;
	return aa->key - bb->key;
}

int main()
{
    int n;
	int head,addr;
	fin>>n>>head;
	map<int,Node> m;		//构建以Address为key的索引集合
	vector<Node> v;
	Node* node = new Node[n];
	int i;
	for(i=0;i<n;i++)
	{
		fin>>addr>>node[i].key>>node[i].next;
		m[addr]=node[i];
	}
	int findKey = head;
	while(findKey != -1)			//根据索引值寻找链 
	{
		map<int,Node>::const_iterator map_it = m.find(findKey);
		if(map_it != m.end())
		{
			Node newNode(findKey, map_it->second.key);
			v.push_back(newNode);
			findKey = map_it->second.next;
		}else
		{
			break;
		}
	}
	
	int size = v.size();
	if(size==0)
	{
		cout<<0<<' '<<-1<<endl;
		system( "PAUSE");
		return 0;
	}
	//sort(v.begin(),v.end());
	qsort(& *v.begin(),size,sizeof(Node),cmp);
	cout<<size<<" "<<setw(5)<<setfill('0')<<v.begin()->next<<endl;
	
	for(i=0;i<size;i++)
	{
		cout<<setw(5)<<setfill('0')<<v[i].next<<' '<<v[i].key<<' ';
		if(i != size-1)
		{
			cout<<setw(5)<<setfill('0')<<v[i+1].next<<endl;
		}else
		{
			cout<<-1<<endl;
		}
	}

    system( "PAUSE");
    return 0;
}