1133. Splitting A Linked List (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (<= 10⁵) which is the total number of nodes, and a positive K (<=1000). 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 [-10⁵, 10⁵], 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 1023333 10 2777700000 0 9999900100 18 1230968237 -6 2333333218 -4 0000048652 -2 -199999 5 6823727777 11 4865212309 7 33218

Sample Output:

33218 -4 6823768237 -6 4865248652 -2 1230912309 7 0000000000 0 9999999999 5 2333323333 10 0010000100 18 2777727777 11 -1

实力不够

实力不够

实力不够

考试的时候在想怎么对结构体排序 写出来什么玩意。。。

问了同学 3次遍历链表即可.。。。

#include<stdio.h>#include<vector>#include<algorithm>using namespace std;struct node{int address;int data;int next;}stu[100000];vector<node>a;int k;int main(){int start,n,i,address,data,next;scanf("%d %d %d",&start,&n,&k);for(i=0;i<n;i++){scanf("%d %d %d",&address,&data,&next);stu[address].address=address;stu[address].data=data;stu[address].next=next;}int p=start;while(p!=-1){data=stu[p].data;if(data<0){a.push_back(stu[p]);}p=stu[p].next;}p=start;while(p!=-1){data=stu[p].data;if(data>=0&&data<=k){a.push_back(stu[p]);}p=stu[p].next;}p=start;while(p!=-1){data=stu[p].data;if(data>k){a.push_back(stu[p]);}p=stu[p].next;}for(i=0;i<a.size()-1;i++){printf("%05d %d %05d\n",a[i].address,a[i].data,a[i+1].address);}printf("%05d %d -1",a[i].address,a[i].data);}