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…

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…

  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…

题目连接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=47066 利用链表换位置时间复杂度为1的优越性,同时也考虑到使用实际的链表对一个数字进行定位需要扫一遍,时间复杂度难以承受,因此使用数组模拟双向链表. 易错点:1.要对特殊位置进行处理,例如xy相邻的情况 2.注意链表头和尾可能在任意一个操作中变化,要进行检查 #include <bits/stdc++.h> using namespace std; ; type…

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…

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…