这道题目的解决方案是双向链表,数据结构本身并不复杂,但对于四种情况的处理不够细致,主要体现在以下几点: 分类讨论不全面,没有考虑特殊情况(本身不需要操作,需要互换的两元素相邻) 没有考虑状态4改变后对其他操作的影响 没有灵活运用数学知识(求偶只需要全部减去奇数即可) 以下贴出AC代码 #include <cstdio>#include <algorithm>const int maxn = 100000 + 10;int left[maxn];int right[maxn];int…

12657 - Boxes in a Line You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4kinds of commands:• 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y )• 2 X Y : move box X…

 省赛B题....手写链表..其实很简单的.... 比赛时太急了,各种手残....没搞出来....要不然就有金了...注:对相邻的元素需要特判..... Problem B Boxes in a Line You have n boxes in a line on the table numbered 1~n from left to right. Your task is to simulate 4 kinds of commands: 1 X Y: move box X to the lef…

Boxes in a Line You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4 kinds of commands: ? 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y ) ? 2 X Y : move box X to th…

  You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4kinds of commands: • 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y ) • 2 X Y : move box X to the right to Y (i…

You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4 kinds of commands: • 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y ) • 2 X Y : move box X to the right to Y (ig…

Boxes in a Line You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4 kinds of commands: • 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y ) • 2 X Y : move box X to th…

  You have n boxes in a line on the table numbered 1...n from left to right. Your task is to simulate 4 kinds of commands: • 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y ) • 2 X Y : move box X to the right to Y (igno…

题目链接:https://www.luogu.org/problemnew/show/UVA12657 分析: 此题使用手写链表+模拟即可.(其实可以用list,而且更简便,但是会大大的超时) 肯定是不能直接用数组模拟了,因为n,m的大小会达到100000. 然后, 1.可以编写一些辅助函数来设置链接关系. 2.注意 op==3的时候,要对xy相邻的情况进行特判,因为有这种情况 2 1 (头节点) 3 1 2 (尾节点) 3.我们会发现如果反转两次,就相当于没有翻转.如果翻转一次,op=1变为o…

题目大意 你有一行盒子,从左到右依次编号为1, 2, 3,…, n.你可以执行四种指令: 1 X Y表示把盒子X移动到盒子Y左边(如果X已经在Y的左边则忽略此指令).2 X Y表示把盒子X移动到盒子Y右边(如果X已经在Y的右边则忽略此指令).3 X Y表示交换盒子X和Y的位置.4 表示反转整条链. 盒子个数n和指令条数m(1<=n,m<=100,000) 题解 用数组来模拟链表操作,对于每个节点设置一个前驱和后继. 1操作是把x的前驱节点和x的后继节点连接,y节点的前驱和x节点连接,x节点和y…