PAT甲级 1133

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1133 Splitting A Linked List （25 分）

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 (≤103). 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 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

题意：给出一串数字，要你根据规则对其重新排列，具体规则是：负数要在正数前面；同时给出一个正数K，要求[0,K]之间的数字要在大于K的数字之前输出，且元素顺序不能改变。输入的第一行：第一个节点的地址节点的数量正数K，接下来N行：节点地址节点值下一个结点地址；输出时的格式：节点地址节点值下一个节点的地址。

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
//保存节点的值和下一个节点的地址
struct node{
int data,next;

}lis[100001];
vector<int> v[3];
int main()
{
    int start,n,k,a;
    cin>>start>>n>>k;
    for(int i=0;i<n;i++){
            //节点的地址做list的下标
cin>>a;
        cin>>lis[a].data>>lis[a].next;
    }
    int p=start;
    while(p!=-1){
        int data=lis[p].data;
        if(data<0)
            v[0].push_back(p);
        else if(data>=0&&data<=k)
            v[1].push_back(p);
        else v[2].push_back(p);
        p=lis[p].next;
    }
    //f标记地址是否为空
    int f=0;
    for(int i=0;i<3;i++){
        for(int j=0;j<v[i].size();j++){
            if(f==0){
                printf("%05d %d",v[i][j],lis[v[i][j]].data);
                f=1;
            }
            else{
                printf(" %05d\n%05d %d",v[i][j],v[i][j],lis[v[i][j]].data);
            }
        }
    }
    cout<<" -1"<<endl;
    return 0;
}