pat02-线性结构1. Reversing Linked List (25)

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

 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 struct node{
     int now,next,data;
 };
 node mem[];
 vector<node> v;
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int h,now,data,next,num,k,temp;
     scanf("%d %d %d",&h,&num,&k);
     int i;
     for(i=;i<num;i++){
         scanf("%d %d %d",&now,&data,&next);
         mem[now].now=now;
         mem[now].data=data;
         mem[now].next=next;
     }
     now=h;
     while(now!=-){
         v.push_back(mem[now]);
         now=mem[now].next;
     }
     int length=v.size();
     int round=length/k;
     for(i=;i<round;i++){
         reverse(v.begin()+i*k,v.begin()+i*k+k);
     }
     for(i=;i<length-;i++){
         printf("%05d %d %05d\n",v[i].now,v[i].data,v[i+].now);
     }
     printf("%05d %d %d\n",v[i].now,v[i].data,-);//注意全反的情况
     return ;
 }