hdu 3436 Queue-jumpers(splay tree)

Queue-jumpers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1225 Accepted Submission(s): 273



Problem Description

Ponyo and Garfield are waiting outside the box-office for their favorite movie. Because queuing is so boring, that they want to play a game to kill the time. The game is called “Queue-jumpers”. Suppose that there are N people numbered from 1 to N stand in a
line initially. Each time you should simulate one of the following operations:

1. Top x :Take person x to the front of the queue

2. Query x: calculate the current position of person x

3. Rank x: calculate the current person at position x

Where x is in [1, N].

Ponyo is so clever that she plays the game very well while Garfield has no idea. Garfield is now turning to you for help.

Input

In the first line there is an integer T, indicates the number of test cases.(T<=50)

In each case, the first line contains two integers N(1<=N<=10^8), Q(1<=Q<=10^5). Then there are Q lines, each line contain an operation as said above.

Output

For each test case, output “Case d:“ at first line where d is the case number counted from one, then for each “Query x” operation ,output the current position of person x at a line, for each “Rank x” operation, output the current person at position x at a line.

Sample Input

3
9 5
Top 1
Rank 3
Top 7
Rank 6
Rank 8
6 2
Top 4
Top 5
7 4
Top 5
Top 2
Query 1
Rank 6

Sample Output

Case 1:
3
5
8
Case 2:
Case 3:
3
6

Author

wzc1989

Source

2010 ACM-ICPC Multi-University
Training Contest（1）——Host by FZU

Recommend

zhouzeyong

题目思路：splay,也可以用线段树或树状数组，主要难点在离散化，这个离散化是把点和点这间的区间都离散化，离散化后的点要保存原来区间的长度和起始点等信息，这样就可用splay模拟了，top操作可以先删点再插入，也可以将要删的点伸展到根结点，然后将它的左子树接到它后继的左边，这样它便成为整棵树最左边的结点了。由于人的编号和结点的index是一一对应的，不会改变，所以可以把这个关系存起来，这样访问的时候就可以直接通过人的编号找到结点了。

splay tree 参考：/article/8983258.html
树状数组参考：http://www.cppblog.com/Yuan/archive/2010/08/18/123871.html