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POJ 2828

中间迷途了好久，渐渐走回正轨。 这道题卡的很死，用g++会TLE但是c++就A了。 思路是利用splay tree，不过公认好方法应该是线段树，用splay耗时3000ms左右，线段树仅2000ms #include <iostream> #include <algorithm> #include <queue> #include <string> #include <vector> #include <cstdi

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springBoot输出类似于c: \Users\ATFWUS\AppData\Local(Temp\tomcat-docbase.613353598438846626.8802\articleImage\2828/B8/28 C：\Users\ATFWUS\AppData\Local(Temp\tomcat-docbase.613353598438846626.8802\articleImage\2828/B8/28 的路径解决办法 第一步 第二部 改变工作

【线段树】poj 2828 Buy Tickets

Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 28643   Accepted: 13533 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

poj 2828 buy Tickets 用线段树模拟带插入的队列

Buy Tickets Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2795 Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a

POJ 2828 Buy Tickets (线段树+反向思维)

这是一道很好的题 开始想到链表，但是在第 \(k\) 个位置插入却很烦人，因为链表查询时 \(O(n)\) 的 因为是线段树题，就想到线段树上去 正着好像很难，那么反着可以么？ 反着，对于每一个操作，就相当于查找前面空出 \(a\) 个格子的位置是哪个，直接线段树上搞一下就行了 正难则反的思维还是很精妙