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

   這道題 真的是 神迷，不知道到底哪裡減了 一分， 看到別人的部落格 說有可能這 n個不是一條鏈上的， 一口老血。

真的太坑了，我的變數都是n.

#include<bits/stdc++.h>
using namespace std;

typedef long long LL;
#define rep(i,a,b) for(int i=a;i<b;++i)

const int N=1e5+10;

map<int,int> pos;

struct Node{
	int p1,p2,val;
}p[N],res[N];

int order[N];

/*
00000 0 0
00000 18 00001
00001 -1 -1

*/

int main(){
	//int n;
	//scanf("%d",&n);
	int st,n,K;
	scanf("%d %d %d",&st,&n,&K);
	rep(i,1,n+1){
		scanf("%d %d %d",&p[i].p1,&p[i].val,&p[i].p2);
		pos[p[i].p1]=i;
	}

	int tmp=st,cnt=0;
	while(n&&tmp!=-1){
		order[cnt++]=pos[tmp];
		tmp=p[pos[tmp]].p2;
	}

	//rep(i,0,cnt)printf("i:%d %d\n",i,order[i]);
	int tot=cnt;
	cnt=1;
	for(int i=0;i<tot;i++){
		if(p[order[i]].val<0){
			res[cnt++]=p[order[i]];
			res[cnt-2].p2=res[cnt-1].p1;
		}
	}

	for(int i=0;i<tot;i++){
		if(p[order[i]].val>=0&&p[order[i]].val<=K){
			res[cnt++]=p[order[i]];
			res[cnt-2].p2=res[cnt-1].p1;
		}
	}

	for(int i=0;i<tot;i++){
		if(p[order[i]].val>K){
			res[cnt++]=p[order[i]];
			res[cnt-2].p2=res[cnt-1].p1;
		}
	}

	//printf("*****\n");
	for(int i=1;i<cnt;i++){//這裡的 cnt 不要寫成 n
		if(i<cnt-1)
			printf("%05d %d %05d\n",res[i].p1,res[i].val,res[i].p2);
		else
			printf("%05d %d -1\n",res[i].p1,res[i].val);
	}

	return 0;
}