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PAT_A1133#Splitting A Linked List

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Source:

PAT A1133 Splitting A Linked List (25 分)

Description:

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 (≤) which is the total number of nodes, and a positive K (≤). The address of a node is a 5-digit nonnegative integer, and NULL is represented by −.

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 [, 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

Keys:

Attention:

Code:

错的

 1 /*
 2 Data: 2019-06-24 16:20:53
 3 Problem: PAT_A1133#Splitting A Linked List
 4 AC:
 5 
 6 题目大意:
 7 重排单链表,不改变原先顺序的前提下,使得
 8 1.负数在前,正数在后;
 9 2.正数中,小于K元素在前,大于K的在后;
10 输入:
11 第一行给出,首元素地址,结点数N<=1e5,阈值K
12 接下来N行,地址,数值,下一个地址
13 输出:
14 地址,数值,下一个地址
15 
16 基本思路:
17 建立数值->地址,地址->数值,地址->下个地址的映射
18 按顺序存储链表中有效的数值,再按要求排序,输出;
19 */
20 
21 #include<cstdio>
22 #include<map>
23 #include<vector>
24 using namespace std;
25 
26 int main()
27 {
28 #ifdef    ONLINE_JUDGE
29 #else
30     freopen("Test.txt", "r", stdin);
31 #endif
32 
33     int first,n,k,address,data,next;
34     vector<int> link,L1,L2;
35     map<int,int> adr,key,nex;
36     scanf("%d%d%d", &first,&n,&k);
37     for(int i=0; i<n; i++)
38     {
39         scanf("%d%d%d", &address,&data,&next);
40         adr[data] = address;
41         key[address] = data;
42         nex[address] = next;
43     }
44     while(first != -1)
45     {
46         if(key[first] < 0)
47             link.push_back(key[first]);
48         else if(key[first] <= k)
49             L1.push_back(key[first]);
50         else
51             L2.push_back(key[first]);
52         first = nex[first];
53     }
54     link.insert(link.end(),L1.begin(),L1.end());
55     link.insert(link.end(),L2.begin(),L2.end());
56 
57     for(int i=0; i<link.size(); i++)
58     {
59         if(i != 0)
60             printf("%05d\n", adr[link[i]]);
61         printf("%05d %d ", adr[link[i]], link[i]);
62     }
63     printf("-1\n");
64 
65     return 0;
66 }

对的

 1 /*
 2 Data: 2019-06-24 16:20:53
 3 Problem: PAT_A1133#Splitting A Linked List
 4 AC: 58:37
 5 
 6 题目大意:
 7 重排单链表,不改变原先顺序的前提下,使得
 8 1.负数在前,正数在后;
 9 2.正数中,小于K元素在前,大于K的在后;
10 输入:
11 第一行给出,首元素地址,结点数N<=1e5,阈值K
12 接下来N行,地址,数值,下一个地址
13 输出:
14 地址,数值,下一个地址
15 
16 基本思路:
17 结构体存储链表信息
18 按顺序存储链表中有效的数值,再按要求排序,输出;
19 */
20 
21 #include<cstdio>
22 #include<map>
23 #include<vector>
24 using namespace std;
25 const int M=1e6+10;
26 struct node
27 {
28     int address,data,next;
29 }info[M];
30 
31 int main()
32 {
33 #ifdef    ONLINE_JUDGE
34 #else
35     freopen("Test.txt", "r", stdin);
36 #endif
37 
38     int first,n,k,adr;
39     vector<node> link,L1,L2;
40     scanf("%d%d%d", &first,&n,&k);
41     for(int i=0; i<n; i++)
42     {
43         scanf("%d", &adr);
44         info[adr].address=adr;
45         scanf("%d%d", &info[adr].data,&info[adr].next);
46     }
47     while(first != -1)
48     {
49         if(info[first].data < 0)
50             link.push_back(info[first]);
51         else if(info[first].data <= k)
52             L1.push_back(info[first]);
53         else
54             L2.push_back(info[first]);
55         first = info[first].next;
56     }
57     link.insert(link.end(),L1.begin(),L1.end());
58     link.insert(link.end(),L2.begin(),L2.end());
59 
60     for(int i=0; i<link.size(); i++)
61     {
62         if(i != 0)
63             printf("%05d\n", link[i].address);
64         printf("%05d %d ", link[i].address, link[i].data);
65     }
66     printf("-1\n");
67 
68     return 0;
69 }

 

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来源: https://www.cnblogs.com/blue-lin/p/11078360.html