02-线性结构3 Reversing Linked List (25分)
- 时间限制：400ms
- 内存限制：64MB
- 代码长度限制：16kB
- 判题程序：系统默认
- 作者：陈越
- 单位：浙江大学
https://pta.patest.cn/pta/test/3512/exam/4/question/62614
Given a constant KKK and a singly linked list LLL, you are supposed to reverse the links of every KKK elements on LLL. For example, given LLL being 1→2→3→4→5→6, if K=3K = 3K=3, then you must output 3→2→1→6→5→4; if K=4K = 4K=4, you must output 4→3→2→1→5→6.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive NNN (≤105le 10^5≤105) which is the total number of nodes, and a positive KKK (≤Nle N≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then NNN lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is an integer, and Next is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
Sample Output:
00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 struct Node 5 { 6 int data; 7 int adr,next; 8 }t; 9 int N,head,K; 10 map<int,Node> Input;//用map储存输入序列，地址作为键值，方便通过地址（head，next）找到相关点 11 stack<Node> Sta,T; 12 vector<Node> Result(N);//储存结果链表 13 vector<Node>::iterator it; 14 int main() 15 { 16 /*输入*/ 17 scanf("%d %d %d",&head,&N,&K); 18 for(int i=0;i<N;i++) 19 { 20 scanf("%d %d %d",&t.adr,&t.data,&t.next); 21 Input[t.adr]=t; 22 } 23 /*处理*/ 24 int num=0; 25 t=Input.find(head)->second;//找到第一个点 26 while(num++<N)//对n个元素入栈处理 27 { 28 int flag=t.next; 29 Sta.push(t); 30 if(t.next!=-1) 31 t=Input.find(t.next)->second; 32 if(num%K==0)//满K个元素，逆序出栈放入Result 33 { 34 while(!Sta.empty()) 35 { 36 Result.push_back(Sta.top()); 37 Sta.pop(); 38 } 39 if(flag==-1)//最后一个结点刚好组成最后K个，结束 40 break; 41 continue; 42 } 43 else if(flag==-1)//最后一个结点多余，将栈中的点出，入，出栈后恢复正序放入Result，结束 44 { 45 while(!Sta.empty()) 46 { 47 T.push(Sta.top()); 48 Sta.pop(); 49 } 50 while(!T.empty()) 51 { 52 Result.push_back(T.top()); 53 T.pop(); 54 } 55 break; 56 } 57 } 58 /*输出*/ 59 it=Result.begin(); 60 printf("%05d %d ",it->adr,it->data); 61 it++; 62 for(;it!=Result.end();it++) 63 { 64 printf("%05d %05d %d ",it->adr,it->adr,it->data); 65 } 66 printf("-1 "); 67 Input.clear(); 68 Result.clear(); 69 return 0; 70 }