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  • PAT1052:Linked List Sorting

    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 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

    思路
    链表排序。
    1.先储存所有输入节点到一个dic中(map模拟)。
    2.根据给定的头结点遍历链表并将访问到的节点插入一个新的序列(vector)中。这个过程能筛掉dic中不在链表中的节点。
    3.对新序列根据key值排序。
    4.按照新排列的顺序修改每个节点指向的next地址
    5.输出:
    1)注意地址格式为标准5位数
    2)如果新序列为空,输出"0 -1"。
    3)最后末尾节点的next地址直接输出-1,而不是5位的格式"-00001"。
    代码
    #include<vector>
    #include<algorithm>
    #include<iostream>
    #include<map>
    using namespace std;
    class node
    {
    public:
       int address;
       int key;
       int next;
    };
    map<int,node> dic;
    vector<node> nodes;
    bool cmp(const node& a,const node& b)
    {
        return a.key < b.key;
    }
    
    int main()
    {
       int N,head;
       while(cin >> N >> head)
       {
           for(int i = 0;i < N;i++)
           {
               int ad;
               cin >> ad;
               dic[ad].address = ad;
               cin >> dic[ad].key >> dic[ad].next;
           }
           while(head != -1)
           {
               nodes.push_back(dic[head]);
               head = dic[head].next;
           }
           sort(nodes.begin(),nodes.end(),cmp);
           int len = nodes.size();
           if(len == 0)
           {
               cout << "0 -1" << endl;
               continue;
           }
           for(int i = 1;i < len;i++)
           {
               nodes[i - 1].next = nodes[i].address;
           }
           nodes[len - 1].next = -1;
           head = nodes[0].address;
           //output
           printf("%d %05d
    ",len,head);
           for(int i = 0;i < len;i++)
           {
               if(i == len - 1)
                 printf("%05d %d -1
    ",nodes[i].address,nodes[i].key);
               else
                 printf("%05d %d %05d
    ",nodes[i].address,nodes[i].key,nodes[i].next);
           }
       }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/0kk470/p/8134366.html
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