线段树 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, h…

Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 12930   Accepted: 6412 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…

Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 22097   Accepted: 10834 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 w…

题目链接:http://poj.org/problem?id=2828 由于最后一个人的位置一定是不会变的,所以我们倒着做,先插入最后一个人. 我们每次处理的时候,由于已经知道了这个人的位置k,这个位置表明,在他之前一定有k个空位,于是将他插在第k+1个位置上.我们可以在线段树上直接二分,根据这个位置,假如这个位置在左子树上,那么一直向左走,否则要减掉左子树剩下的位置后再向右边走,同时更新沿途非叶节点的空闲位置数量(减一),手工画一下图就知道了. /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏…

[poj2828]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…

http://poj.org/problem?id=2828 Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 10478   Accepted: 5079 Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a lo…

Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 19725   Accepted: 9756 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…

Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 16607   Accepted: 8275 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…

https://vjudge.net/problem/POJ-2828 题目意思:有n个数,进行n次操作,每次操作有两个数pos, ans.pos的意思是把ans放到第pos 位置的后面,pos后面的数就往后推一位.最后输出每个位置的ans. 思路:根据题 目可知,最后插入的位置的数才是最终不变的数,所以可以从最后的输入作第1个放入,依此类推,倒插入.在插入时也有一定的技术,首先创建一棵空线段树时,每个节点记录当前范围内有多少个空位置.在插入时,要注意,一个数放入之后那么这个位置就不用管了,那么…

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…