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…
[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…
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…
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! ┓┏┓┏…
题意  n个人排队  每一个人都有个属性值  依次输入n个pos[i]  val[i]  表示第i个人直接插到当前第pos[i]个人后面  他的属性值为val[i]  要求最后依次输出队中各个人的属性值 从头到尾看的话  队列是动态的   无法操作  可是反过来看时  pos[i]就能够表示第i个人前面有多少个空位了  然后想到了用线段树做就简单了  线段树维护相应区间还有多少个空位  每次把i放到前面刚好有pos[i]个空位的位置即可了  详细看代码 #include <cstdio> #d…
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: 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…
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…
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…