02-线性结构2. Reversing Linked List (25)

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

反转节点；我要学一下printf，scanf。。。dyx~
#include<iostream>#include<algorithm>#include<vector>#include<iomanip>using namespace std;const int Max=100009;class Node{    public:    int add;    int data;    int  Next;};Node node[Max];int main(){    int firstadd,num,k;    cin>>firstadd>>num>>k;//k是反转节点;    //输入数据;    for(int i=1;i<=num;i++)    {        Node wyx;        cin>>wyx.add>>wyx.data>>wyx.Next;        node[wyx.add]=wyx;    }    //建立向量；    vector <Node> dyx;    //将数据压入向量当中;    int first=firstadd;    while(first!=-1)    {        dyx.push_back(node[first]);        first=node[first].Next;    }    int len=dyx.size();    int rev=len/k;//看有几节反转点；    for(int i=1;i<=rev;i++)    {        int st=(i-1)*k;        int end=i*k;        reverse(dyx.begin()+st,dyx.begin()+end);    }        for(int i=0;i<len-1;i++)        {            cout<<setw(5)<<setfill('0')<<dyx[i].add<<" ";            cout<<dyx[i].data<<" ";            cout<<setw(5)<<setfill('0')<<dyx[i+1].add<<endl;        }        cout<<setw(5)<<setfill('0')<<dyx[len-1].add<<" ";        cout<<dyx[len-1].data<<" ";        cout<<-1<<endl;        return 0;}