Problem B Boxes in a Line
手写链表。。其实很简单的。。。。
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 left to Y (ignore this if X is already the left of Y)
- 2 X Y: move box X to the right to Y (ignore this if X is already the right of Y)
- 3 X Y: swap box X and Y
- 4: reverse the whole line.
Commands are guaranteed to be valid, i.e. X will be not equal to Y.
For example, if n=6, after executing 1 1 4, the line becomes 2 3 1 4 5 6. Then after executing 2 3 5, the line becomes 2 1 4 5 3 6. Then after executing 3 1 6, the line becomes 2 6 4 5 3 1. Then after executing 4, then line becomes 1 3 5 4 6 2
Input
There will be at most 10 test cases. Each test case begins with a line containing 2 integers n, m(1<=n, m<=100,000). Each of the following m lines contain a command.
Output
For each test case, print the sum of numbers at odd-indexed positions. Positions are numbered 1 to n from left to right.
Sample Input
6 4
1 1 4
2 3 5
3 1 6
4
6 3
1 1 4
2 3 5
3 1 6
100000 1
4
Output for the Sample Input
Case 1: 12
Case 2: 9
Case 3: 2500050000
The Ninth Hunan Collegiate Programming Contest (2013)
Problemsetter: Rujia Liu
Special Thanks: Feng Chen, Md. Mahbubul Hasan
#include <iostream>
#include <cstdio>
#include <cstring> #define toleft p
#define toright p^1 using namespace std; const int INF=0x3f3f3f3f;
typedef long long int LL; int lian[][],n,m,p; void Init()
{
p=;
for(int i=;i<=n;i++)
{
lian[i][]=i-;
lian[i][]=i+;
}
lian[][toleft]=INF;
lian[][toright]=;
lian[n+][toleft]=n;
lian[n+][toright]=INF;
} void cmdfour()
{
p=p^;
} void cmdone(int x,int y)
{
int xl=lian[x][toleft],xr=lian[x][toright];
int yl=lian[y][toleft],yr=lian[y][toright]; if(xr==y) return; lian[xl][toright]=xr;
lian[xr][toleft]=xl; lian[x][toright]=y;
lian[x][toleft]=yl; lian[yl][toright]=x;
lian[y][toleft]=x;
} void cmdtwo(int x,int y)
{
int xl=lian[x][toleft],xr=lian[x][toright];
int yl=lian[y][toleft],yr=lian[y][toright]; if(xl==y) return; lian[xl][toright]=xr;
lian[xr][toleft]=xl; lian[x][toleft]=y;
lian[x][toright]=yr; lian[y][toright]=x;
lian[yr][toleft]=x;
} void cmdthree(int x,int y)
{
int xl=lian[x][toleft],xr=lian[x][toright];
int yl=lian[y][toleft],yr=lian[y][toright]; if(xr!=y&&xl!=y)
{
lian[x][toleft]=yl;
lian[x][toright]=yr;
lian[yl][toright]=x;
lian[yr][toleft]=x; lian[y][toleft]=xl;
lian[y][toright]=xr;
lian[xl][toright]=y;
lian[xr][toleft]=y;
}
else
{
if(xr==y)
{
lian[x][toright]=yr;
lian[x][toleft]=y; lian[y][toright]=x;
lian[y][toleft]=xl; lian[xl][toright]=y;
lian[yr][toleft]=x;
}
else if(xl==y)
{
lian[x][toleft]=yl;
lian[x][toright]=y; lian[y][toleft]=x;
lian[y][toright]=xr; lian[xr][toleft]=y;
lian[yl][toright]=x;
} }
} LL getSum()
{
LL ans=;
int cnt=,pos=;
if(n%==||p==)
{
if(p) p=;
while(true)
{
if(cnt&) ans+=(LL)lian[pos][toright];
pos=lian[pos][toright];
cnt++;
if(cnt>=n+) break;
}
}
else if(p==)
{
if(p) p=;
while(true)
{
if(cnt%==) ans+=(LL)lian[pos][toright];
pos=lian[pos][toright];
cnt++;
if(cnt>=n+) break;
}
}
return ans;
} int main()
{
int cas=,cmd,x,y;
while(scanf("%d%d",&n,&m)!=EOF)
{
Init(); while(m--)
{
scanf("%d",&cmd);
if(cmd==)
{
scanf("%d%d",&x,&y);
cmdone(x,y);
}
else if(cmd==)
{
scanf("%d%d",&x,&y);
cmdtwo(x,y);
}
else if(cmd==)
{
scanf("%d%d",&x,&y);
cmdthree(x,y);
}
else if(cmd==)
{
cmdfour();
}
}
LL ans=getSum();
printf("Case %d: %lld\n",cas++,ans);
}
return ;
}
Problem B Boxes in a Line的更多相关文章
- Boxes in a Line
Boxes in a Line You have n boxes in a line on the table numbered 1 . . . n from left to right. Your ...
- uva-12657 - Boxes in a Line(双向链表)
12657 - Boxes in a Line You have n boxes in a line on the table numbered 1 . . . n from left to righ ...
- Boxes in a Line(移动盒子)
You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to sim ...
- C - Boxes in a Line 数组模拟链表
You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simul ...
- UVa 12657 Boxes in a Line(应用双链表)
Boxes in a Line You have n boxes in a line on the table numbered 1 . . . n from left to right. Your ...
- Boxes in a Line UVA - 12657
You have n boxes in a line on the table numbered 1...n from left to right. Your task is to simulat ...
- UVA 12657 Boxes in a Line 双向链表
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 ...
- Boxes in a Line UVA - 12657 (双向链表)
题目链接:https://vjudge.net/problem/UVA-12657 题目大意:输入n,m 代表有n个盒子 每个盒子最开始按1~n排成一行 m个操作, 1 x y :把盒子x放到y ...
- 2019 SCUT SE 新生训练第四波 L - Boxes in a Line——双向链表
先上一波题目 https://vjudge.net/contest/338760#problem/L 这道题我们维护一个双向链表 操作1 2 3 都是双向链表的基本操作 4操作考虑到手动将链表反转时间 ...
随机推荐
- Mac 配置 vim
Mac 配置 vim 安装 vim 7.4 cd ~ hg clone https://code.google.com/p/vim/ cd vim ./configure --with-feature ...
- 使用IntelliJ IDEA和Maven构建Java web项目并打包部署
爱编程爱分享,原创文章,转载请注明出处,谢谢! http://www.cnblogs.com/fozero/p/6120375.html 一.背景 现在越来越多的人使用IntelliJ IDEA工具进 ...
- CF 204B Little Elephant and Cards
题目链接: 传送门 Little Elephant and Cards time limit per test:2 second memory limit per test:256 megab ...
- python 培训之爬虫
1. 输入文件为 fufang_list.txt yaofang_a aaiwan 阿艾丸 yaofang_a aaiwulingsan 阿艾五苓散 yaofang_a acaitang 阿菜汤 ya ...
- jQuery基础(1) -- jQuery 语法
通过 jQuery,您可以选取(查询,query) HTML 元素,并对它们执行"操作"(actions).jQuery 语法jQuery 语法是通过选取 HTML 元素,并对选取 ...
- BZOJ2286: [Sdoi2011]消耗战
建出虚树dp. 把询问点按dfs序排序,用一个以dfs序为关键字的单调栈(以深度为关键字也是一样的),每次将一个询问点与栈顶的点的lca入栈,再将这个询问点入栈,在这个过程中建出一棵树就是虚树.具体看 ...
- windows系统下安装MySQL
可以运行在本地windows版本的MySQL数据库程 序自从3.21版以后已经可以从MySQL AB公司获得,而且 MYSQL每日的下载百分比非常大.这部分描述在windows上安装MySQL的过程. ...
- AngularJs:String类型和JSON相互转换
最近一周做了一个页面,制作的过程中遇到各种问题,从中可以看出本人的js基础还不够扎实,angularjs也只是刚入门的水平,现在将制作过程中遇到的问题一一汇总,方便以后查阅. 一.String类型和J ...
- python学习笔记-(一)初识python
1.python的前世今生 想要充分的了解一个人,无外乎首先充分了解他的过去和现在:咱们学习语言也是一样的套路 1.1 python的历史 Python(英国发音:/ˈpaɪθən/ 美国发音:/ˈp ...
- select()函数以及FD_ZERO、FD_SET、FD_CLR、FD_ISSET
http://hi.baidu.com/%B1%D5%C4%BF%B3%C9%B7%F0/blog/item/e7284ef16bcec3c70a46e05e.html select函数用于在非阻塞中 ...