1052. Linked List Sorting (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 0000111111 100 -100001 0 2222233333 100000 1111112345 -1 3333322222 1000 12345
Sample Output:
5 1234512345 -1 0000100001 0 1111111111 100 2222222222 1000 3333333333 100000 -1
首先给出N个地址，再给出第一个地址
接着每个地址  地址存放值  下一个地址
……
目的：通过第一个地址获得这条所带有的全部存储单元，在把这些单元通过存储的值从小到大排序。输出地址注意格式
用到
#include<vector>动态数组#include<algorithm>排序#include<iomanip>设置格式

评测结果
时间结果得分题目语言用时(ms)内存(kB)用户8月12日 15:04答案正确251052C++ (g++ 4.7.2)2793752datrilla
测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确130815/151答案正确13083/32答案正确13083/33答案正确27937523/34答案正确13081/1

#include<iostream> #include<vector>#include<algorithm>#include<iomanip>#define MAX 100000using namespace std; struct Links{  int key;  int next;  Links(){}  Links(int k,int n):key(k),next(n){}};void readln(Links*H, int N){  Links temp;  int addr;  while (N--)  {    cin >> addr >> temp.key >> temp.next;    H[addr] = temp;  }}void Together(Links*H, int Headaddr, vector< Links>*sortH){   while (Headaddr != -1)  {    sortH->push_back(Links(H[Headaddr].key, Headaddr));    Headaddr = H[Headaddr].next;   }}bool Hcmp(const Links &A, const Links &B){  return A.key < B.key;}void format(int a, int b, int c){  cout << setw(5) << setfill('0') << a <<" "<< b << " "<<setw(5) << setfill('0') << c << endl;}int main(){     Links H[MAX];  int N, Headaddr;    vector< Links>sortH;  cin >> N >> Headaddr;  readln(H, N);  Together(H, Headaddr, &sortH);  sort(sortH.begin(), sortH.end(), Hcmp);  Headaddr=sortH.size();  if (Headaddr == 0)  {    cout << "0 -1" << endl;  }  else  {    cout << Headaddr-- << " " << setw(5) << setfill('0') << sortH[0].next << endl;    for (N = 0; N < Headaddr; N++)      format(sortH[N].next, sortH[N].key,  sortH[N + 1].next);     cout << setw(5) << setfill('0')<< sortH[N].next << " " << sortH[N].key << " -1" << endl;  }  system("pause");  return 0;}