错误的代码：

#include <cstdio>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <string>
using namespace std;struct Node
{int Nodenow;int Nodenext;int data;Node *next;
};struct store
{int now;int next;int data;
}a[100005],b[100005];int turn[100005];void Node_Print(Node *head)
{Node *p = head;while(p != NULL){if(p -> Nodenext != -1)printf("%05d %d %05d",p->Nodenow,p -> data,p -> Nodenext);else{printf("%05d %d %d",p->Nodenow,p -> data,p -> Nodenext);}p = p -> next;if(p != NULL)printf("\n");}
}Node * ReversePart_Node(Node *head,int start,int end)
{if(start == end)return head;Node *pPrv = NULL;Node *pNow = head;Node *pNext = NULL;Node *BeforeStart = NULL; //开始的结点前的结点 Node *StartNode = NULL; //开始的时候的结点int i,j;//for循环指向需要反转的位置 for(i = 1; i <= start; i++){if(i == start){BeforeStart = pPrv;StartNode = pNow;}pPrv = pNow;pNow = pNow -> next;pNext = pNow -> next;}for(i = 1; i <= end - start; i++)  {pNext = pNow -> next; //保存下一个 pNow -> next = pPrv; //改变指向 pPrv = pNow; //移动 pPrv和 pNow pNow = pNext;}//反转部分 未反转前的第一个结点 反转之后变成该反转部分的最后一个结点 StartNode -> next = pNow; //连接 if(BeforeStart == NULL) //从head开始反转 {head = pPrv; //pNow 反转过后是 该反转部分的第一个结点 }else {BeforeStart -> next = pPrv; //连接反转部分前面的结点 和 该反转部分 }return head;
}int inturn(int ibegin,int stot)
{int i,j,found;int tot = 0;int cnt = 1;if(!turn[ibegin]) return 0;b[cnt++] = a[turn[ibegin]];found = a[turn[ibegin]].next;while(found != -1){b[cnt++] = a[turn[found]];found = a[turn[found]].next;}cnt--;return cnt;
}Node *Node_Creat(int tot)
{Node *head;head = (Node *)malloc(sizeof(Node));if(head == NULL){printf("Overflow\n");exit(1);}head -> data = b[1].data;head -> Nodenow = b[1].now;head -> Nodenext = b[1].next;Node *p1,*p2;p1 = p2 = head;for(int i = 2; i <= tot; i++){p1 = (Node *)malloc(sizeof(Node));if(p1 == NULL){printf("Overflow\n");exit(1);}p1 -> data = b[i].data;p1 -> Nodenow = b[i].now;p1 -> Nodenext = b[i].next;p2 -> next = p1;p2 = p1;}p2 -> next = NULL;return head;
}int main()
{int stot,reverse;int ibegin;int i,j;cin >> ibegin >> stot >> reverse;int tot = 0; //结点数 memset(turn,0,sizeof(turn));for(i = 1; i <= stot; i++){cin >> a[i].now >> a[i].data >> a[i].next;turn[a[i].now] = i;}tot = inturn(ibegin,stot);Node *head;head = Node_Creat(tot);for(i = 1; i <= tot/reverse; i++){head = ReversePart_Node(head, (i - 1)*reverse + 1, i*reverse);}head = changedata(head);Node_Print(head);return 0;
}
/*
00100 12 13
72222 12 -1
62222 11 72222
52222 10 62222
42222 9 52222
32222 8 42222
22222 7 32222
00000 4 99999
00100 1 12309
68237 6 22222
33218 3 00000
99999 5 68237
12309 2 33218
*/

后来根据一些数据的对比，发现反转后的结点下一个地址要发生改变，于是添加了一个函数改变结点中地址的值。

Node *changedata(Node *head)
{Node *p = head;while(p -> next != NULL){p -> Nodenext = p -> next -> Nodenow;p = p -> next;}p -> Nodenext = -1;return head;
}

改了一下输入方式，但是还是超时(挠头