【PAT A1052】链表题 Linked List Sorting

Linked List Sorting
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.
题目大意：给出N个结点的地址，数据，指针域，链表的首地址，设计一个程序把这个链表按照数据从大到小排序，输出链表个数，首地址，各个结点的地址，数据，指针域。（最重要的就是要考虑有无效结点和没有结点吧） 还是考虑用静态链表，因为简单，遍历输入结点，有效结点flag赋值true并记录个数count，排序算法中以flag为依据，对flag从大到小排序，前面的就都是有效结点了，都是有效结点再对数据从小到大排序。再根据之前统计的count输出。

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=100005;
struct NODE{
	int address,data,next;
	bool flag;
}node[maxn];
bool cmp(NODE a,NODE b){
	if(a.flag==false||b.flag==false){
		return a.flag>b.flag;
	}else{
		return a.data<b.data;
	}
}
int main(void){
	for(int i=0;i<maxn;i++){
		node[i].flag=false;
	}
	int n,begin,address,data,next;
	scanf("%d %d",&n,&begin);
	for(int i=0;i<n;i++){
		scanf("%d %d %d",&address,&data,&next);
		node[address].address=address;
		node[address].data=data;
		node[address].next=next;
	}
	int count=0,p=begin;
	while(p!=-1){
		node[p].flag=true;
		count++;
		p=node[p].next;
	}
	if(count==0){
		printf("0 -1\n");
	}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;
}