02 linear structure 3 reversing linked list

02- Linear structure 3 Reversing Linked List

fraction 25

author Chen Yue

Company Zhejiang University

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

 

analysis ：
    For the former rev-1 Inverted linked list , The next address of the last node of each inverted linked list is the next inverted linked list
    First address ;
    about res Inverted linked list , If n%k==0, Then the next address of the last node is -1, Otherwise, the next node of the last node
    The address is the first address that cannot form an inverted linked list .

/**
     analysis ：
     For the former rev-1 Inverted linked list , The next address of the last node of each inverted linked list is the next inverted linked list 
     First address ;
     about res Inverted linked list , If n%k==0, Then the next address of the last node is -1, Otherwise, the next node of the last node 
     The address is the first address that cannot form an inverted linked list .
*/

#include <iostream>
#include <algorithm>
using namespace std;

const int maxn=1e5+10;
struct Node
{
    int addr,data,next;
    int flag;
    Node()
    {
        flag=2*maxn;
    }
};
bool cmp(Node &a,Node &b)
{
    return a.flag<b.flag;
}
int main()
{
    int head,n,k;
    scanf("%d%d%d",&head,&n,&k);
    Node node[maxn];
//    for(int i=0;i<maxn;++i)
//        node[i].flag=0;
    int addr;
    for(int i=0;i<n;++i)
    {
        scanf("%d",&addr);
        scanf("%d%d",&node[addr].data,&node[addr].next);
        node[addr].addr=addr;
    }

    int i=0;
    while(head!=-1)
    {
        node[head].flag=++i;
        head=node[head].next;
    }
    n=i;    // There are redundant nodes that are not on the linked list 
    sort(node,node+maxn,cmp);
    /**
    cout << "jidu\n";
    for(int i=0;i<n;++i)
        printf("%05d %d %05d\n",node[i].addr,node[i].data,node[i].next);
    cout << "lo\n";
    */
    int rev=n/k;
    for(int i=0;i<rev;++i)
    {
        for(int j=k-1;j>=0;--j)
        {
            if(i!=rev-1)  // If it is the last time of reversing the linked list 
            {
                int index=i*k+j;
                if(j==0)    // If it is the last node of each trip 
                    printf("%05d %d %05d\n",node[index].addr,node[index].data,node[(i+2)*k-1].addr);
                else
                    printf("%05d %d %05d\n",node[index].addr,node[index].data,node[index-1].addr);
            }
            else
            {
                int index=i*k+j;
                if(j==0)
                {
                    if(n%k==0)
                        printf("%05d %d -1\n",node[index].addr,node[index].data);
                    else
                        printf("%05d %d %05d\n",node[index].addr,node[index].data,node[(i+1)*k].addr);
                }
                else
                    printf("%05d %d %05d\n",node[index].addr,node[index].data,node[index-1].addr);
            }
        }
    }

    for(int i=rev*k;i<n;++i)    // The remaining nodes that cannot form an inverted linked list 
    {
        if(i==n-1)
            printf("%05d %d -1\n",node[i].addr,node[i].data);
        else
            printf("%05d %d %05d\n",node[i].addr,node[i].data,node[i+1].addr);
    }
    return 0;
}