csu 1365 双向链表模拟超时
1365: Play with Chain
Time Limit: 5 Sec Memory Limit: 128 MBSubmit: 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 ath diamond to the
bth diamond. And then insert it after the cth 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 5th
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 ath diamond to
the bth 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 ith number is the number of the ith
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 }
