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, 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

解题思路：

可能有多个链表干扰，这时候可以用hash来查找在同一个链表里面的节点。

#include"iostream"#include"stdio.h"#include"stdlib.h"#include"string.h"#include"algorithm"#include"vector"using namespace std;#define MAX 100000#define LEN 6typedef struct node{char addr[LEN];int d;char next[LEN];}Node;int main(){long N,K;char start[LEN];scanf("%s%ld%ld",start,&N,&K);Node b[MAX];/*题目的内存还是很大的，空间换时间*/vector<Node> a;for(int i=0;i<N;i++){Node n;scanf("%s%d%s",n.addr,&n.d,n.next);long index= atol(n.addr);b[index] = n;}/*可能有多个链表,过滤掉不属于start开头链表的噪声节点*/long index2 = atol(start);while(index2!=-1){a.push_back(b[index2]);index2 = atol(b[index2].next);}long size = a.size();long len = size/K;for(long i=1;i<=len;i++){    long start = (i-1)*K;    long end = i*K;    reverse(a.begin()+start,a.begin()+end);    }   for(long i=0;i<size-1;i++){printf("%s %d %s\n",a[i].addr,a[i].d,a[i+1].addr);}printf("%s %d %d\n",a[size-1].addr,a[size-1].d,(-1));    return 0;  }