02-线性结构1. Reversing Linked List (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
8000 B
判题程序
Standard
作者
CHEN, Yue
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 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 N (<= 105) which is the total number of nodes, and a positive K (<=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 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, 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
提交代码
1 #include <cstdio>
2 #include <cstring>
3 #include <string>
4 #include <queue>
5 #include <stack>
6 #include <cmath>
7 #include <iostream>
8 #include <algorithm>
9 using namespace std;
10 struct node{
11 int now,next,data;
12 };
13 node mem[100005];
14 vector<node> v;
15 int main(){
16 //freopen("D:\\INPUT.txt","r",stdin);
17 int h,now,data,next,num,k,temp;
18 scanf("%d %d %d",&h,&num,&k);
19 int i;
20 for(i=0;i<num;i++){
21 scanf("%d %d %d",&now,&data,&next);
22 mem[now].now=now;
23 mem[now].data=data;
24 mem[now].next=next;
25 }
26 now=h;
27 while(now!=-1){
28 v.push_back(mem[now]);
29 now=mem[now].next;
30 }
31 int length=v.size();
32 int round=length/k;
33 for(i=0;i<round;i++){
34 reverse(v.begin()+i*k,v.begin()+i*k+k);
35 }
36 for(i=0;i<length-1;i++){
37 printf("%05d %d %05d\n",v[i].now,v[i].data,v[i+1].now);
38 }
39 printf("%05d %d %d\n",v[i].now,v[i].data,-1);//注意全反的情况
40 return 0;
41 }