PAT (Advanced Level) Practise 1074 Reversing Linked List (25)

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

题意：给你一个链表的头，让你每隔k个节点倒置一下节点

解题思路：模拟，这题好像比较坑的一点是不一定是一个链表

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;struct node{    int val,nt;}a[100009];int s,n,k,x,ans[100009],cnt;int main(){    while(~scanf("%d%d%d",&s,&n,&k))    {        for(int i=1;i<=n;i++) scanf("%d",&x),scanf("%d%d",&a[x].val,&a[x].nt);        cnt=0;        for(int i=s;~i;i=a[i].nt) ans[cnt++]=i;        for(int i=0;i+k<=cnt;i+=k)        {            for(int j=0;j<=k;j++)            {                if(i+j>=i+k-j-1) break;                swap(ans[i+j],ans[i+k-j-1]);            }        }        for(int i=0;i<cnt;i++)        {            printf("%05d %d ",ans[i],a[ans[i]].val);            if(i==cnt-1) printf("-1\n");            else printf("%05d\n",ans[i+1]);        }    }    return 0;}