csu 1365 双向链表模拟超时

1365: Play with Chain

Time Limit: 5 Sec  Memory Limit: 128 MB
Submit:
21  Solved: 5
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Description

YaoYao is fond of playing his chains. He has
a chain containing n diamonds on it. Diamonds are numbered from 1 to n.
At
first, the diamonds on the chain is a sequence: 1, 2, 3, …, n.
He will
perform two types of operations:

CUT a b c: He
will first cut down the chain from the a^th diamond to the
b^th diamond. And then insert it after the c^th diamond on
the remaining chain.
For example, if n=8, the chain is: 1 2 3 4 5 6 7 8; We
perform “CUT 3 5 4”, Then we first cut down 3 4 5, and the remaining chain would
be: 1 2 6 7 8. Then we insert “3 4 5” into the chain before 5^th
diamond, the chain turns out to be: 1 2 6 7 3 4 5 8.

FLIP a
b: We first cut down the chain from the a^th diamond to
the b^th diamond. Then reverse the chain and put them back to the
original position.
For example, if we perform “FLIP 2 6” on the chain: 1 2 6
7 3 4 5 8. The chain will turn out to be: 1 4 3 7 6 2 5 8

He wants to
know what the chain looks like after perform m operations. Could you help him?

Input

There will be multiple test cases in a test
data.
For each test case, the first line contains two numbers: n and m (1≤n,
m≤3*100000), indicating the total number of diamonds on the chain and the number
of operations respectively.
Then m lines follow, each line contains one
operation. The command is like this:

CUT a b c   // Means a CUT operation, 1 ≤
a ≤ b ≤ n, 0≤ c ≤ n-(b-a+1).
FLIP a b        // Means a FLIP operation, 1 ≤ a
< b ≤ n.

The input ends up with two negative
numbers, which should not be processed as a case.

Output

For each test case, you should print a line
with n numbers. The i^th number is the number of the i^th
diamond on the chain.

Sample Input

8 2
CUT 3 5 4
FLIP 2 6
-1 -1

Sample Output

1 4 3 7 6 2 5 8

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 using namespace std;

 struct node
 {
     int num;
     struct node *next;
     struct node *father;
 };
 struct node *head;
 void mem(struct node *p)
 {
     p->num=;
     p->next=NULL;
     p->father=NULL;
 }
 void CUT(int l,int r,int k,int n)
 {
     if(k==l-)return;
     if(l==&&r==n)return ;
     struct node *p,*q,*st,*ed,*hxl;
     int i,cur;
     if(k<l) cur=r;
     else
     {
         cur=r-l++k;
         k=cur;
     }
     p=head;
     for(i=;i<=cur&&p!=NULL;i++)
     {
         p=p->next;
         if(i==l) st=p;
         if(i==r) ed=p;
         if(i==k) hxl=p;
     }
     if(k==) hxl=head;
     p=st->father;
     q=ed->next;
     p->next=q;
     if(q!=NULL) q->father=p;

     p=hxl->next;
     ed->next=hxl->next;
     if(p!=NULL) p->father=ed;

     hxl->next=st;
     st->father=hxl;
 }
 void FLIP(int l,int r)
 {
     int i,tmp,tom;
     struct node *st,*ed,*q;
     q=head;
     for(i=;i<=r;i++)
     {
         q=q->next;
         if(i==l)    st=q;
         if(i==r)    ed=q;
     }
     tom=(r-l+)/;
     while(tom--)
     {
         tmp=st->num;
         st->num=ed->num;
         ed->num=tmp;

         st=st->next;
         ed=ed->father;
     }
 }
 int main()
 {
     int n,m,i;
     int l,r,k;
     char cur[];
     struct node *p,*q;
     while(scanf("%d%d",&n,&m)>)
     {
         if(n==-&&m==-)break;
         head=(struct node*)malloc(sizeof(struct node));
         mem(head);
         p=head;
         for(i=;i<=n;i++)
         {
             q=(struct node*)malloc(sizeof(struct node));
             q->num=i;
             q->next=p->next;
             p->next=q;
             q->father=p;
             p=q;
         }
         getchar();
         while(m--)
         {
             scanf("%s",cur);
             if(cur[]=='C')
             {
                 scanf("%d%d%d",&l,&r,&k);
                 CUT(l,r,k,n);
             }
             else if(cur[]=='F')
             {
                 scanf("%d%d",&l,&r);
                 FLIP(l,r);
             }
         }
         p=head;
         for(i=;i<=n;i++)
         {
             q=p;
             p=p->next;
             free(q);
             printf("%d",p->num);
             if(i!=n)printf(" ");
             else printf("\n");
         }
         free(p);
     }
     return ;
 }