1074. Reversing Linked List (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1

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提交代码

http://www.patest.cn/contests/pat-a-practise/1074

#include <stdio.h>#include <iostream>#include <algorithm>#include <vector>#include <set>#include <string>#include <queue> using namespace std; #define N 100005 #define INF 1 << 30 struct node{  int addr ;  int data ;  int next  ;};node vn[N] ;int main(){  //freopen("in.txt" , "r" , stdin) ;  int firstAdd , n  , K ;  scanf("%d%d%d",&firstAdd , &n , &K ) ;  int i ;  int Address , Data , Next ;  node nt ;  for(i = 0 ;i < n ; i++)  {    scanf("%d%d%d" , &Address , &Data , &Next) ;    nt.addr = Address ;    nt.data = Data ;    nt.next = Next ;    vn[Address] = nt ;  }  vector<node> vv ;  while(firstAdd != -1)  {    vv.push_back(vn[firstAdd]) ;    firstAdd = vn[firstAdd].next ;  }  int len = vv.size() ;  int count =  len / K ;  vector<node> ans ;  for(i = 0 ; i < count ; i++)  {    int last = (i+1)*K - 1 ;    int first = i * K ;    for(int j = last ; j >= first; j --)    {      ans.push_back(vv[j]) ;    }  }  for(i = count*K ; i < len ; i++)  {    ans.push_back(vv[i]) ;  }  printf("%05d %d " , ans[0].addr , ans[0].data ) ;  for(i = 1 ;i < len ; i++)  {    printf("%05d\n%05d %d " , ans[i].addr , ans[i].addr , ans[i].data) ;  }  printf("-1\n") ;    return 0;}