C++PATA1052

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 (<10 5 ) 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 [−10
5 ,10 5 ], 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

核心思路

就是给静态链表排排序，用vector拿到地址，用地址给data进行排序，拍完序最后进行输出

完整代码

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
const int maxn = 1000010;
struct Node{
    int data;
    int next;
}node[maxn];
bool cmp(int x,int y){
    return node[x].data<node[y].data;
}
int main()
{
    int s1,n;//s1,s2分别代表首地址
    scanf("%d%d",&n,&s1);
    int address,next;//结点地址与后缀结点地址
    char data;
    int k;
    while(n--){
        cin >> k >> node[k].data >> node[k].next;
    }
    int p;
    vector<int>v;
    for(p=s1;p!=-1;p=node[p].next){
        v.emplace_back(p);
    }
    sort(v.begin(),v.end(),cmp);

    cout << v.size();
    int i;
    for(i=0;i<v.size();i++){
        printf(" %05d\n%05d %d",v[i],v[i],node[v[i]].data);
    }
    cout << " -1";
    return 0;
}

方法2，遍历链表设置为true

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
const int maxn = 100005;
struct Node {
    int address,data,next;
    bool flag; // 结点是否在链表上
}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中有个无效结点，就把它放在后面
    }else{
        return a.data < b.data; //如果都是有效结点，则按要求排序
    }
}

int main()
{
    for(int i = 0;i<maxn;i++){
        node[i].flag = false;
    }
    int n,begin,address;
    scanf("%d%d",&n,&begin);
    for(int i =0;i<n;i++){
        scanf("%d",&address);
        scanf("%d%d",&node[address].data,&node[address].next);
        node[address].address = address;
    }
    int count = 0,p = begin;//标记链表，对flag进行标记，同时计数有效结点个数(步骤3)
    while(p != -1){
        node[p].flag = true;
        count ++;
        p = node[p].next;
    }
    if(count == 0){
        printf("0 -1");
    }else{
        sort(node,node+maxn,cmp);
        printf("%d %05d\n",count,node[0].address);
        for(int i =0;i<count;i++){
            if(i!= count-1){
                printf("%05d %d %05d\n",node[i].address,node[i].data,node[i+1].address);
            }else{
                printf("%05d %d -1\n",node[i].address,node[i].data);
            }
        }
    }
    return 0;
}