PAT甲级——A1052 Linked List Sorting

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

Then

Address Key Next

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

Output Specification:

For each test case, the output format is the same as that of the input, where

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

注意点
（1）题目给出的结点中可能有不在链表中的无效结点

（2）最后一个测试点测试的是首地址为 - 1，即为空链表的情况，此时只输出0 - 1

（3）输出时结点地址除 - 1外要有5位数字，不够则在高位补0 。所以地址 - 1要进行特判输出

 1 #include <iostream>
 2 #include <map>
 3 using namespace std;
 4 int main()
 5 {
 6     int N, head, maxV = -1000000;
 7     cin >> N >> head;
 8     map<int, int>value;
 9     map<int, pair<int, int>>data;
10     for (int i = 0; i < N; ++i)
11     {
12         int addr, val, next;
13         cin >> addr >> val >> next;
14         data.insert(make_pair(addr, make_pair(val, next)));
15     }
16     if (data.find(head) == data.end())
17     {
18         cout << 0 << " " << -1 << endl;
19         return 0;//该链表为空
20     }
21     while (head != -1)//遍历一下数据，找到是链表中的数据
22     {
23         value.insert(make_pair(data[head].first, head));
24         head = data[head].second;
25     }
26     printf("%d %05d\n", value.size(), value.begin()->second);
27     for (auto ptr = value.begin(); ptr != value.end(); ++ptr)
28     {
29         if (ptr == value.begin())
30             printf("%05d %d ", ptr->second, ptr->first);
31         else
32             printf("%05d\n%05d %d ", ptr->second, ptr->second, ptr->first);
33     }
34     printf("%d\n", -1);
35     return 0;
36 }