A1052.Linked List Sorting

最新推荐文章于 2020-07-05 14:52:32 发布

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本文介绍了一种基于输入的链表结构进行排序的算法实现。该算法通过读取一系列包含地址、键值和指向下一个节点地址的数据，构建起一个链表，并最终按照键值的升序重新排列这些结构。此外，文章还提供了完整的C++代码示例。

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

Code:

#include "cstdio"
#include "algorithm"

using namespace std;

const int maxn = 100010;
struct NODE
{
    int address,key,next;
    bool flag;
} node[maxn];

bool cmp(NODE a,NODE b)
{
    if(a.flag == false || b.flag == false)
    {
        return a.flag > b.flag; //只要a和b中有一个无效结点，就吧它放到后面
    }
    else
    {
        return a.key < b.key; //只要都是有效结点，则按从小到大排序
    }
}

int main()
{
    for(int i = 0; i < maxn; i++)  //初始化将flag=false
    {
        node[i].flag = false;
    }

    int n,s;
    scanf("%d%d",&n,&s);
    int address,key,next;
    for(int i = 0; i < n; i++)
    {
        scanf("%d%d%d",&address,&key,&next);
        node[address].key = key;
        node[address].next = next;
        node[address].address = address;
    }

    int c = 0,p = s;
    while(p != -1)  //枚举链表，对flag标记同时计数有效结点个数
    {
        node[p].flag = true;
        c++;
        p = node[p].next;
    }
    if(c == 0) //特例，链表没有有效结点
    {
        printf("0 -1");
    }
    else
    {
        sort(node, node + maxn, cmp);
        printf("%d %05d\n",c,node[0].address);
        for(int i = 0; i < c; i++)
        {
            if(i != c - 1)
            {
                printf("%05d %d %05d\n",node[i].address,node[i].key,node[i+1].address);
            }
            else
            {
                printf("%05d %d -1\n",node[i].address,node[i].key);
            }
        }
    }
    return 0;
}