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 4 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218
Sample Output:
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1
简单的模拟题,以空间换时间：
#include <cstdio> #include <vector> #include <algorithm> using namespace std;const int NUM=100001;struct Node {int address;int data;int next; };Node nodes[NUM]; vector<Node> list; int main() {int fnAddress,N,K;scanf("%d %d %d",&fnAddress,&N,&K);int i;for(i=0;i<N;++i){Node node;scanf("%d %d %d",&node.address,&node.data,&node.next);nodes[node.address]=node;}int address=fnAddress;while(address!=-1)//去噪 {list.push_back(nodes[address]);address=nodes[address].next;}int size=list.size();int round=size/K;int start,end;for(i=1;i<=round;++i){start=(i-1)*K;end=i*K;reverse(list.begin()+start,list.begin()+end);}for(i=0;i<size-1;++i){printf("%.5d %d %.5d\n",list[i].address,list[i].data,list[i+1].address);}printf("%.5d %d %d\n",list[i].address,list[i].data,-1);return 0; }
