1133 Splitting A Linked List-PAT甲级

Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10​5​​) which is the total number of nodes, and a positive K (≤10​3​​). The address of a node is a 5-digit nonnegative integer, and NULL is represented by −1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer in [−10​5​​,10​5​​], and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218

Sample Output:

33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1

本题主要考察的是静态链表的复杂应用， 选取两个指标作为排序的依据，一个是结点值所在的区间，另一个是结点所在链表上的编号，在这里需要注意的是即便给出的样例中有些结点可能只是形式，不在链表上，因此在排序的过程中，因保证其在链表上在排序，id值的大小可以明确结点是否在链表上

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=100010;
const int inf=0x3f3f3f3f;
struct Node{
	int id;//用来将同一类型按照原链表输出，指定它们在链表中出现的顺序 
	int address;
	int data;
	int next;
	int flag;//用来记录数字大小的标志 
	bool operator < (const Node &a)const{
		if(flag!=a.flag&&id!=inf&&a.id!=inf){//负数 不大于k的数 大于k的数 
			return flag<a.flag;
		}else{
			return id<a.id;
		}
	}
}node[N];
int first,n,k;
int main(){
	//初始化结点在链表中的顺序
	for(int i=0;i<N;i++){
		node[i].id=inf;
		node[i].flag=inf;
	} 
	scanf("%d%d%d",&first,&n,&k);
	for(int i=1;i<=n;i++){
		int ad;
		scanf("%d",&ad);
		scanf("%d%d",&node[ad].data,&node[ad].next);
		node[ad].address=ad;
		if(node[ad].data<0){
			node[ad].flag=0;
		}else if(node[ad].data<=k){
			node[ad].flag=1;
		}else{
			node[ad].flag=2;
		}
	}
	int p=first,cnt=0;
	while(p!=-1){
		node[p].id=cnt;
		cnt++;
		p=node[p].next;
	}
	sort(node,node+N);
	for(int i=0;i<cnt;i++){
		if(i==cnt-1){
			printf("%05d %d -1\n",node[i].address,node[i].data);
			break;
		}
		printf("%05d %d %05d\n",node[i].address,node[i].data,node[i+1].address);
	}
	return 0;
}