1074. Reversing Linked List

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

天啊，看错题了。。。我以为是反转前K个结点。。。找了好长时间错都没看出来！！！一定注意审题！！！！！

和之前一个题一样，有无关的节点，所以要注意输入的长度L和真实链长是不同的。

。。。。。。清醒点

#include<iostream>#include<algorithm>#include<vector>using namespace std;typedef struct node{int value;int next;};node p[100000];int main(){int root,l,k;cin>>root>>l>>k;vector<int>list;vector<int>val;for(int i=0;i<l;++i){int id,value,next;cin>>id>>value>>next;p[id].next=next;p[id].value=value;}int tmp=root;while(tmp!=-1){list.push_back(tmp);val.push_back(p[tmp].value);tmp=p[tmp].next;}l=list.size();int pos=0;while(pos+k<=l){reverse(list.begin()+pos,list.begin()+k+pos);reverse(val.begin()+pos,val.begin()+k+pos);pos+=k;}for(int i=0;i<l-1;++i)printf("%05d %d %05d\n",list[i],val[i],list[i+1]);printf("%05d %d -1\n",list[l-1],val[l-1]); return 0;}