1052 Linked List Sorting (25分)

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本文介绍了一种使用链表进行排序的算法实现，通过利用map数据结构的特性，将链表节点按照其key值进行排序。文章详细解释了输入输出规格，并提供了一个样例输入输出，最后给出了完整的C++代码实现。

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：按照链表的数据域排序后输出链表。

思路：因为key值是唯一的，可以利用map自带的排序功能，map的key保存链表的key，map的value保存链表的address，这样直接输出就好了，注意输出的格式为5位字符～

#include<iostream>
#include<vector>
#include<algorithm>
#include<map>
using namespace std;

int main(){

    int n, start;
    cin >> n >> start;
    int data[100010] = {0};
    int order[100010] = {0};
    map<int, int> m;
    for(int i = 0; i < n; i++){
        int add, key, next;
        scanf("%d %d %d", &add, &key, &next);
        data[add] = key;
        order[add] = next;
        
    }
    for(int i = start; i != -1; i = order[i]){
        m[data[i]] = i;
    }
    
    map<int, int>::iterator it = m.begin();
    if((m.size()) == 0){
        printf("0 -1\n");
        return 0;
    }else
    {
        printf("%d %05d\n", int(m.size()), it -> second);
    }
    
    
    while (it != m.end())
    {
        printf("%05d %d", it -> second, it -> first);
        it++;
        if(it != m.end()){
            printf(" %05d\n", it -> second);
        }else
        {
            printf(" -1\n");
        }
        
        
    }
    


    return 0;



}