PAT 1052 Linked List Sorting [一般]

1052 Linked List Sorting （25 分）

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

 题目大意：给出一个链表，按照其中存储的key值大小重新排序。

最终AC:

#include <iostream>
#include <algorithm>
#include<cstdio>
#include <vector>
using namespace std;

struct Node{
    int now,num,next;
}node[100010];
vector<Node> vt;
bool cmp(Node& a,Node& b){
    return a.num<b.num;
}
int main()
{
    int n,f;
    scanf("%d%d",&n,&f);
    int t,num,next;
    for(int i=0;i<n;i++){
        cin>>t>>num>>next;
        node[t].num=num;
        node[t].next=next;
        node[t].now=t;
    }
    for(int index=f;index!=-1;index=node[index].next){
        vt.push_back(node[index]);
    }

    //有一个疑问就是如果输入数据为0 0 ，怎么输出？
    if(vt.size()==0){
        printf("0 -1");
    }else{
        sort(vt.begin(),vt.end(),cmp);
        printf("%d %05d
",vt.size(),vt[0].now);
    for(int i=0;i<vt.size();i++){
        if(i!=vt.size()-1){
            printf("%05d %d %05d
",vt[i].now,vt[i].num,vt[i+1].now);
        }else{
            printf("%05d %d -1",vt[i].now,vt[i].num);
        }
    }
    }
    return 0;
}

//自己遇到了很多坑。。。

1.对于输入是0 0不知道如何输出，看了别人的发现是输出0和-1，记住吧，以后对于链表长度为0的，就是输出数量0，然后直接就是结束-1.

2.第一次提交时有段错误，感觉到是数组定义长度的问题，一开始定义长度为100000，后来又加了10可以了，最后一个测试点，那么以后在定义时就比平常多加10就可以啦

3.之后提交发现第4个测试点过不去，这是为什么呢？去牛客网上发现是如下错误：

测试用例:
1 00000
00000 5 -1

对应输出应该为:

1 00000
00000 5 -1

你的输出为:

1 0

最终在输出起点坐标时加上了%05d，就可以了。。。。