PAT 1052. Linked List Sorting (25) 【分3种情况讨论】

1052. Linked List Sorting (25)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

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

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提交代码

用Hash存储结构体的思想 去 模拟链表。

要点：

（1）并不是所有输入的节点都是链表上的节点；

（2）若链表为空，输出0 -1.

#include<stdio.h>#include<algorithm>using namespace std;const int MAX=100010;struct E{int addr;int num;int next;   //下标只能采用Hash思想，才能接得住。。}r[MAX],ans[MAX];  //输入的数据可能有的并不在链表中，需要除掉！bool cmp(E p,E q){return p.num<q.num;}int main(){intn,m,i,a,b,c;    //freopen("G:\\in.txt","r",stdin);while(scanf("%d%d",&n,&m)!=EOF){for(i=0;i<n;i++){scanf("%d%d%d",&a,&b,&c);r[a].addr=a;r[a].num=b;r[a].next=c;}int tmp=m,cnt=0;   //首地址做第一个下标while(tmp!=-1){ans[cnt++]=r[tmp];tmp=r[tmp].next;}sort(ans,ans+cnt,cmp);          if(cnt>1){                  printf("%d %05d\n",cnt,ans[0].addr); //格式控制。。for(i=0;i<cnt-1;i++)printf("%05d %d %05d\n",ans[i].addr,ans[i].num,ans[i+1].addr);                  printf("%05d %d -1\n",ans[cnt-1].addr,ans[cnt-1].num);                }         if(cnt==1){printf("%d %05d\n",cnt,ans[0].addr);                  printf("%05d %d -1\n",ans[0].addr,ans[0].num);                }          if(cnt==0)                  printf("0 -1\n");}    return 0;}