题目传送门 Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 22611   Accepted: 11045 Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue… The Lunar New…

Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue… The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. Now, he ha…

Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 19012   Accepted: 9442 Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue… The Lunar New Year wa…

题目链接:poj 2828 Buy Tickets 题目大意:给定N,表示有个人,给定每一个人站入的位置,以及这个人的权值,如今按队列的顺序输出每一个人的权值. 解题思路:第K大元素,非常巧妙,将人入队的顺序倒过来看,就是纯第K大问题,然后用树状数组还是线段树就都能够做了. C++ 线段树 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int ma…

前提是数的范围较小 1 数据范围:O(n) 2 查第k大的数i:log(n)(树状数组查询小于等于i的数目)*log(n)(二分找到i) 3 添加:log(n) (树状数组) 4 删除:log(n) (树状数组) 团体程序设计天梯赛 L3-002. 堆栈 /*数据范围:O(n) 查第k大的数i:log(n)(树状数组查询小于等于i的数目)*log(n)(二分找到i) 添加:log(n) (树状数组) 删除:log(n) (树状数组) */ #include <cstdio> #include…

Buy Tickets Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue… The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there…

题目链接:http://poj.org/problem?id=2828 题意就是给你n个人,然后每个人按顺序插队,问你最终的顺序是怎么样的. 反过来做就很容易了,从最后一个人开始推,最后一个人位置很容易就确定了,那最后第二个人的位置也可以推(与最后一个人的位置无关)...依次就都可以确定所有的人了. 用前缀和的思想,要是这个人的位置确定了,那么就标记这个人位置的值为0,然后回溯更新,跟求逆序对个数的思想比较类似. 线段树: #include <iostream> #include <cs…

描述 有个脑筋急转弯是这样的:有距离很近的一高一低两座桥,两次洪水之后高桥被淹了两次,低桥却只被淹了一次,为什么?答案是:因为低桥太低了,第一次洪水退去之后水位依然在低桥之上,所以不算“淹了两次”.举例说明:假定高桥和低桥的高度分别是5和2,初始水位为1第一次洪水:水位提高到6(两个桥都被淹),退到2(高桥不再被淹,但低桥仍然被淹)第二次洪水:水位提高到8(高桥又被淹了),退到3.没错,文字游戏.关键在于“又”的含义.如果某次洪水退去之后一座桥仍然被淹(即水位不小于桥的高度),那么下次洪水来临水…

题意: 有两种操作:1.在[l,r]上插入一条值为val的线段 2.问p位置上值第k小的线段的值(是否存在) 特别的,询问的时候l和p合起来是一个递增序列 1<=l,r<=1e9:1<=val<=1e6; 1<=k<=1e9 思路: 因为l和p总体是递增的,第i个询问的p一定大于i之前所有操作的l,而前面能影响到i的答案的只有r>=p的线段.由此可以想到将l,r,p合起来离散化,从左往右扫描,遇到l,就在权值线段树上插入对应的val,遇到r就删除对应的val,而遇…

题目链接:http://poj.org/problem?id=2182 Lost Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12736   Accepted: 8168 Description N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, t…