迷之1025 反转链表

#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
*/

Node *changedata(Node *head)
{
Node *p = head;

while(p -> next != NULL)
{
p -> Nodenext = p -> next -> Nodenow;
p = p -> next;
}
p -> Nodenext = -1;

return head;
}