标签：
Time Limit: 20 Sec
Memory Limit: 256 MB
http://acm.hdu.edu.cn/showproblem.php?pid=4286
Input
The first line contains an integer T(1<=T<=10), the number of test cases.
Then T test cases follow. For each test case, the first line contains an integer N(1<=N<=500000), the number of data at the beginning. The second line contains N integers, means the integer in each data, from left to right. The third line contains two integers L and R (1<=L<=R<=N), the positions of hand "L" and hand "R". It means that hand "L" is at the left of the L-th data and hand "R" is at the right of the R-th data. The fourth line contains one integer M(1<=M<=500000), the number of commands. Then M lines follow, each line contains a command in the above format. All the integers in the data will in range [-10000,10000].
It is guaranteed that there are always some data between hand "L" and "R", and if the hand is at the left/right end of the data line, it will not receive the command MoveLeft/MoveRight.
Because of large input, please use scanf instead of cin.
Output
For each test case, output the integers in the data from left to right in one line, separated in a single space.
Because of large output, please use printf instead of cout.
Sample Input
2
5
1 2 3 4 5
1 5
5
MoveLeft R
Insert R 6
Reverse
Delete R
Insert L 7
5
6536 5207 2609 6604 -4046
1 3
5
Delete L
Insert R -9221
Reverse
Delete L
MoveRight L
Sample Output
7 6 4 3 2 5
2609 5207 6604 -4046
题意
翻转，删除，插入操作
题解：
splay或者双端队列，但是我智商太低了，所以splay过不了
代码
#include<cstdio> #include<cstring> using namespace std; struct node { int value; int pre,next; } num[1000005]; int l,ll,r,rr,cnt; void MoveLefe() { char ch[5]; scanf("%s",ch); if(ch[0]==‘L‘) { if(num[ll].pre==l) { l=ll; ll=num[ll].next; } else { l=ll; ll=num[ll].pre; } } else { if(num[r].pre==rr) { rr=r; r=num[r].next; } else { rr=r; r=num[r].pre; } } } void MoveRight() { char ch[5]; scanf("%s",ch); if(ch[0]==‘L‘) { if(num[l].pre==ll) { ll=l; l=num[l].next; } else { ll=l; l=num[l].pre; } } else { if(num[rr].pre==r) { r=rr; rr=num[rr].next; } else { r=rr; rr=num[rr].pre; } } } void Insert() { char ch[5]; int value; scanf("%s%d",ch,&value); if(ch[0]==‘L‘) { num[cnt].pre=ll; num[cnt].next=l; if(num[ll].pre==l) num[ll].pre=cnt; else num[ll].next=cnt; if(num[l].pre==ll) num[l].pre=cnt; else num[l].next=cnt; l=cnt++; num[l].value=value; } else { num[cnt].pre=rr; num[cnt].next=r; if(num[rr].pre==r) num[rr].pre=cnt; else num[rr].next=cnt; if(num[r].pre==rr) num[r].pre=cnt; else num[r].next=cnt; r=cnt++; num[r].value=value; } } void Delete() { char ch[5]; int next; scanf("%s",ch); if(ch[0]==‘L‘) { if(num[l].pre==ll) next=num[l].next; else next=num[l].pre; if(num[ll].pre==l) num[ll].pre=next; else num[ll].next=next; if(num[next].pre==l) num[next].pre=ll; else num[next].next=ll; l=next; } else { if(num[r].pre==rr) next=num[r].next; else next=num[r].pre; if(num[rr].pre==r) num[rr].pre=next; else num[rr].next=next; if(num[next].pre==r) num[next].pre=rr; else num[next].next=rr; r=next; } } void Reverse() { if(num[r].pre==rr) num[r].pre=ll; else num[r].next=ll; if(num[l].pre==ll) num[l].pre=rr; else num[l].next=rr; if(num[ll].pre==l) num[ll].pre=r; else num[ll].next=r; if(num[rr].pre==r) num[rr].pre=l; else num[rr].next=l; l=l^r; r=l^r; l=l^r; } void out(int n) { bool first=true; int pre=0; for(int i=num[0].next;i!=n+1;) { if(first) { printf("%d",num[i].value); first=false; } else printf(" %d",num[i].value); if(num[i].next!=pre) { pre=i; i=num[i].next; } else { pre=i; i=num[i].pre; } } printf("\n"); } int main() { int cas,n,m; char s[20]; scanf("%d",&cas); for(; cas--;) { scanf("%d",&n); cnt=n+2; for(int i=1; i<=n; ++i) { scanf("%d",&num[i].value); num[i].pre=i-1; num[i].next=i+1; } scanf("%d%d",&l,&r); num[0].pre=0; num[0].next=1; num[n+1].pre=n; ll=num[l].pre; rr=num[r].next; scanf("%d",&m); for(; m--;) { scanf("%s",s); if(strcmp(s,"MoveLeft")==0) MoveLefe(); else if(strcmp(s,"MoveRight")==0) MoveRight(); else if(strcmp(s,"Insert")==0) Insert(); else if(strcmp(s,"Delete")==0) Delete(); else if(strcmp(s,"Reverse")==0) Reverse(); } out(n); } return 0; }
HDU 4286 Data Handler 双向链表/Splay
标签：
原文地址：http://www.cnblogs.com/qscqesze/p/4662225.html
