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

 

 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 }

 

posted @ 2015-08-11 13:52  Deribs4  阅读(182)  评论(0)  编辑  收藏  举报