题目链接:hdu 1890 Robotic Sort 题意: 给你n个数,每次找到第i小的数的位置,然后输出这个位置,然后将这个位置前面的数翻转一下,然后删除这个数,这样执行n次. 题解: 典型的splay区间翻转+删点. 我们把数据排序,然后记录一下每个数原来的位置,然后splay建树的时候用原来的位置来对应,这样val[i].second就直接是这个数在splay中的那个节点. (当然你也可以普通建树,然后手动记录位置). 然后我们把要找的那个数对应的节点旋转到根,然后根左边的size+i就…

Robotic Sort Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) [Problem Description] Somewhere deep in the Czech Technical University buildings, there are laboratories for examining mechanical and electrical properti…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1890 Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description Somewhere deep in the Czech Technical University buildings, there are laboratories for examining…

Robotic Sort Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3456    Accepted Submission(s): 1493 Problem Description Somewhere deep in the Czech Technical University buildings, there are labora…

Robotic Sort Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1640    Accepted Submission(s): 711 Problem Description Somewhere deep in the Czech Technical University buildings, there are laborat…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=1890 [题意] 给定一个序列,每次将i..P[i]反转,然后输出P[i],P[i]定义为当前数字i的所在位置.相等的两个数排序后相对位置不变. [思路] 由于相对位置不变,所以可以根据数值与位置重编号. 依旧使用直接定位从上到下旋转至根的splay写法.每次将i结点旋转至根,则答案为左儿子大小+i,然后将i删掉合并左右儿子. 需要注意合并时判断左右儿子是否为空,以及各种pushdown下传标记.…

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1890 如下: #include<cstdio> #include<cstdlib> #include<iostream> #include<algorithm> using std::sort; using std::swap; ; struct node{ int val, pos; }rec[]; inline bool cmp(const node &am…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1890 题解:splay又一高级的功能,区间旋转这个是用线段树这些实现不了的,这题可以学习splay的旋转方法还有splay tree是按照中序来的,也就是说中序遍历后会得到原序列所以建树和线段树差不多稍微有点不一样.其实splay tree核心操作就是splay就是将某个点移到goal下面优化bst的操作. #include <cstdio> #include <iostream> #…

Robotic Sort Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3913    Accepted Submission(s): 1717 Problem Description Somewhere deep in the Czech Technical University buildings, there are labora…

kpm大神说可以用块状链表写...但是我不会...写了个splay.... 先离散化 , 然后splay结点加个min维护最小值 , 就可以了... ( ps BZOJ 3506 题意一样 , 双倍经验 ) ----------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #inclu…