csu 1365 双向链表模拟超时

时间：2014-03-11 01:25:17 收藏：0 阅读：491

1365: Play with Chain

Time Limit: 5 Sec Memory Limit: 128 MB
Submit: 21 Solved: 5
[Submit][Status][Web Board]

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 }

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原文：http://www.cnblogs.com/tom987690183/p/3592313.html