PAT 1025反转链表问题求大神解答

#include <iostream>
#include <string.h>
#include <string>
using namespace std;
struct Node{
    string Address;        //结点地址
    int Data;               //改结点存储的整数数据
    string Next;           //下一个结点的地址
};
int main()
{
    std::ios::sync_with_stdio(false);
    string head;
    int N,K;
    cin >> head >> N >> K;
    struct Node node[N];
    string tempstore;
    int flag = 0;
    for(int i = 0;i < N;i++)
        cin >> node[i].Address >> node[i].Data >> node[i].Next;
    struct Node storeNode[N];
    for(int i = 0;i < N;i++)                //找出首结点存入storeNode
    {
        if(head == node[i].Address)
        {
            storeNode[0] = node[i];
            flag = 1;
            break;
        }
    }
    if(flag == 0)
        return 0;

    for(int i = 0;i < N;i++)                //找出接下来的结点按顺序存入
    {
        flag = 0;
        if(storeNode[i].Next != "-1")
        {

            for(int j = 0;j < N;j++)
            {
                if(storeNode[i].Next == node[j].Address)
                {
                    flag = 1;
                    storeNode[i+1] = node[j];
                    break;
                }
            }
            if(flag == 0)       //如果没有找到下一个结点地址则退出查找
            {
                N = i+1;
                break;
            }
        }
        else{
            N= i+1;
            break;

        }

    }
    if(N >= K)
    {

        struct Node temp;
        int t = N;
        int i = 0,j;
        while(t >= K)
        {
            j = i;
            while(j < (i+K)/2 )
            {
                temp = storeNode[j];
                storeNode[j] = storeNode[i+K-j-1];
                storeNode[i+K-j-1] = temp;
                j++;
            }
            t-=K;
            i+=K;
        }
        int k;
        for(k = 0;k < N-1;k++)
        {
            storeNode[k].Next = storeNode[k+1].Address;
        }
        storeNode[k].Next = "-1";
    }
    for(int i = 0;i < N;++i)
        cout << storeNode[i].Address << ' ' << storeNode[i].Data << ' ' << storeNode[i].Next << endl;
    return 0;
}