PAT 甲级 1041 Linked List Sorting (25)

时间限制 1000 ms内存限制 65536 KB代码长度限制 100 KB判断程序 Standard(来自小小)

题目描述

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.

输入描述:

Each input file contains one test case. For each case, the first line contains a positive N (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 [-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.

输出描述:

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.

输入例子:

5 00001
11111 100 -1
00001 0 22222
33333 100000 11111
12345 -1 33333
22222 1000 12345

输出例子:

5 12345
12345 -1 00001
00001 0 11111
11111 100 22222
22222 1000 33333
33333 100000 -1

用结构体模拟的链表进行排序，数组的下标必须是当前节点的地址。

还有就是排完序之后next的值没有变，下一个节点的地址不再是next了，要输出b[i+1]的地址。

感觉链表题仿佛没有一个是用链表写的。。哈哈哈哈哈，，

#include<bits/stdc++.h>

using namespace std;
const int N=112345;

struct node
{
int ad;
int data;
int next;
}a
,b
;

bool cmp(node a,node b)
{
return a.data<b.data;
}

int main()
{
int n,x;
scanf("%d%d",&n,&x);
for(int i=0;i<=n-1;i++)
{
int ad,data,next;
scanf("%d%d%d",&ad,&data,&next);
a[ad].ad=ad;
a[ad].data=data;
a[ad].next=next;
}
if(x==-1)
{
printf("0 -1\n");
}
else
{
int i=0;
while(x!=-1)
{
b[i].ad=a[x].ad;
b[i].data=a[x].data;
b[i].next=a[x].next;
x=a[x].next;
i++;
}
sort(b,b+i,cmp);
printf("%d %05d\n",i,b[0].ad);
for(int j=0;j<=i-2;j++)
{
printf("%05d %d %05d\n",b[j].ad,b[j].data,b[j+1].ad);
}
printf("%05d %d -1\n",b[i-1].ad,b[i-1].data);
}
return 0;
}