Looploop Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1651    Accepted Submission(s): 517 Problem Description XXX gets a new toy named Looploop. The toy has N elements arranged in a loop, an…

Queue-jumpers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3139    Accepted Submission(s): 848 Problem Description Ponyo and Garfield are waiting outside the box-office for their favorite mo…

http://acm.hdu.edu.cn/showproblem.php?pid=4453 题意:很多种操作:1.add x,将从光标起的 k2 个数全部加上 x:2.reverse,将从光标起的 k1 个数全部反转:3.insert x,在光标处的后一位插入值为 x 的数:4.delete,删除光标所在位置的数:5.move x,如果x是2,将光标右移,否则将光标左移.6.查询光标所在位置的值. 思路:在ACM中第一次写了6000+bytes的代码,把Splay几乎所有操作都汇集了,是一个很…

Looploop Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 781    Accepted Submission(s): 220 Problem Description XXX gets a new toy named Looploop. The toy has N elements arranged in a loop, an a…

给定两种操作,一种是把一个数列的某一段切下来插到剩余数列的某一个位置上. 一种是翻转操作,把数列的某一段进行翻转. 都是Splay的基本操作.标准的Rotateto调整出 [a,b]区间.然后对[a,b]区间修改parent标记和child标记.然后记住PushUp把修改标记推到树根上.简单一点就直接对某个节点spaly(x,0)就OK! 1: #include <cstdio> 2: #include <iostream> 3: #include <vector> 4…

Robotic Sort Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3340    Accepted Submission(s): 1423 Problem Description Somewhere deep in the Czech Technical University buildings, there are labor…

Shaolin Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 3021    Accepted Submission(s): 1273 Problem Description Shaolin temple is very famous for its Kongfu monks.A lot of young men go to Shaol…

Play with Chain Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6779    Accepted Submission(s): 2678 Problem Description YaoYao is fond of playing his chains. He has a chain containing n diamond…

题意:与区间查询点更新,点有20W个,询问区间的最大值.曾经用线段树,1000+ms,今天的伸展树,890没ms,差不多. 第一次学习伸展树,一共花了2个单位时间,感觉伸展树真很有用,也很好玩.现在只学了一点点.切个点更新试试. 大致思路:用编号(数组)作为树的键值建树,每插一个数,沿路节点更新最大值(每个结点有一个附加信息标记以之为子树的树所有点的最大值).所以,查询时[i,j],只要把i-1伸展到树根,把j+1伸展到I-1下面,那么j+1的左子树就是要的区间了!查该子树根值信息即可(特判端点…

约会安排 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 433    Accepted Submission(s): 145 Problem Description 寒假来了,又到了小明和女神们约会的季节. 小明虽为屌丝级码农,但非常活跃,女神们常常在小明网上的大段发言后热情回复“呵呵”,所以,小明的最爱就是和女神们约会.与此同时,也有…