pat1074_备份

1074. Reversing Linked List (25)

时间限制

300 ms

内存限制

32000 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, andNext 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

有点错误，明天再看看

#include <cstdio>#include <list>#include <algorithm>using namespace std ;struct node {int head ,key,end ;bool operator < ( const node & b ) const {return key < b.key ;}};list<node> list2 ;struct node list1[100001] , list3[100001] ;int len3 = 0 , len1;int head , L,K ;void input(){node x ;scanf("%d%d%d", &head ,&L , &K ) ;for ( int i = 0 ; i < L; i++ ){scanf( "%d%d%d", &x.head , &x.key , &x.end  ) ;list1[x.head] = x ;}} void showList2(){  list<node>::iterator it = list2.begin ()  ;  int count = 0 ;  while ( count < list2.size() -1 )  {  printf("%05d %d %05d\n", it->head , it->key , it->end ) ;  count++ ;  it++ ;  }  printf("%05d %d -1", it->head , it->key ) ; }void deal_(){   int h= head ;   int mod , x ;   int loc ; while ( h != -1 ) {list3[len3++] = list1[h ];h= list1[h].end ;} sort( list3, list3+len3) ; x = L/K ;mod = L%K ; for ( int i = 0 ; i < x ; i++ ){for (int j = 0 ; j < K ; j++ ){ loc = K*i+j ; list2.push_back(list3[loc]) ; }list2.reverse() ;}for ( int i = 0 ; i < mod ; i++ ){list2.push_back(list3[loc+i+1]) ;} list<node>::iterator it = list2.begin() ; list<node>::iterator it2 = list2.begin() ; it2++ ; int count = 0 ; while ( count < list2.size()-1  ) { count++ ; it->end = it2->head ; it++ ; it2++ ; }  it->end = -1 ; showList2() ;}int main ( void ){input() ;deal_() ; return 0 ;}