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

  1 #include<iostream>

  2 #include<stdio.h>

  3 #include<cstring>

  4 #include<cstdlib>

  5 #include<algorithm>

  6 using namespace std;

  7 

  8 struct node

  9 {

 10     int num;

 11     struct node *next;

 12     struct node *father;

 13 };

 14 struct node *head;

 15 void mem(struct node *p)

 16 {

 17     p->num=0;

 18     p->next=NULL;

 19     p->father=NULL;

 20 }

 21 void CUT(int l,int r,int k,int n)

 22 {

 23     if(k==l-1)return;

 24     if(l==1&&r==n)return ;

 25     struct node *p,*q,*st,*ed,*hxl;

 26     int i,cur;

 27     if(k<l) cur=r;

 28     else 

 29     {

 30         cur=r-l+1+k;

 31         k=cur;

 32     }

 33     p=head;

 34     for(i=1;i<=cur&&p!=NULL;i++)

 35     {

 36         p=p->next;

 37         if(i==l) st=p;

 38         if(i==r) ed=p;

 39         if(i==k) hxl=p;

 40     }

 41     if(k==0) hxl=head;

 42     p=st->father;

 43     q=ed->next;

 44     p->next=q;

 45     if(q!=NULL) q->father=p;

 46 

 47     p=hxl->next;

 48     ed->next=hxl->next;

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

 50 

 51     hxl->next=st;

 52     st->father=hxl;

 53 }

 54 void FLIP(int l,int r)

 55 {

 56     int i,tmp,tom;

 57     struct node *st,*ed,*q;

 58     q=head;

 59     for(i=1;i<=r;i++)

 60     {

 61         q=q->next;

 62         if(i==l)    st=q;

 63         if(i==r)    ed=q;

 64     }

 65     tom=(r-l+1)/2;

 66     while(tom--)

 67     {

 68         tmp=st->num;

 69         st->num=ed->num;

 70         ed->num=tmp;

 71 

 72         st=st->next;

 73         ed=ed->father;

 74     }

 75 }

 76 int main()

 77 {

 78     int n,m,i;

 79     int l,r,k;

 80     char cur[10];

 81     struct node *p,*q;

 82     while(scanf("%d%d",&n,&m)>0)

 83     {

 84         if(n==-1&&m==-1)break;

 85         head=(struct node*)malloc(sizeof(struct node));

 86         mem(head);

 87         p=head;

 88         for(i=1;i<=n;i++)

 89         {

 90             q=(struct node*)malloc(sizeof(struct node));

 91             q->num=i;

 92             q->next=p->next;

 93             p->next=q;

 94             q->father=p;

 95             p=q;

 96         }

 97         getchar();

 98         while(m--)

 99         {

100             scanf("%s",cur);

101             if(cur[0]=='C')

102             {

103                 scanf("%d%d%d",&l,&r,&k);

104                 CUT(l,r,k,n);

105             }

106             else if(cur[0]=='F')

107             {

108                 scanf("%d%d",&l,&r);

109                 FLIP(l,r);

110             }

111         }

112         p=head;

113         for(i=1;i<=n;i++)

114         {

115             q=p;

116             p=p->next;

117             free(q);

118             printf("%d",p->num);

119             if(i!=n)printf(" ");

120             else printf("\n");

121         }  

122         free(p);

123     }

124     return 0;

125 }