Before the cows head home for rest and recreation, Farmer John wantsthem to get some intellectual stimulation by playing a game.
The game board comprises N (1 <= N <= 15) identical holes in theground, all of which are initially empty. A cow moves by eithercovering exactly one hole with a rock, or uncovering exactly onepreviously covered hole. The game state is defined by which holesare covered with rocks and which aren't. The goal of the game isfor the cows to reach every possible game state exactly once andthen return to the state with all holes uncovered.The cows have been having a tough time winning the game. Below isan example of one of their games:
Holes
time 1 2 3
-----------------
0 O O O Initially all of the holes are empty
1 O O X The cow covers hole 3
2 X O X The cow covers hole 1
3 X O O The cow uncovers hole 3
4 X X O The cow covers hole 2
5 O X O The cow uncovers hole 1
6 O X X The cow covers hole 3
7 X X X The cow covers hole 1
Now the cows are stuck! They must uncover one hole and no matterwhich one they uncover they will reach a state they have alreadyreached. For example if they remove the rock from the second hole they will reach the state (X O X) which they already visited at
time 2.
Below is an example of a valid solution for N=3 holes:
Holes
time 1 2 3
-----------------
0 O O O Initial state: all of the holes are empty
1 O X O The cow covers hole 2
2 O X X The cow covers hole 3
3 O O X The cow uncovers hole 2
4 X O X The cow covers hole 1
5 X X X The cow covers hole 2
6 X X O The cow uncovers hole 3
7 X O O The cow uncovers hole 2
8 O O O The cow uncovers the 1st hole
which returns the game board to the start having, visited each state once.
The cows are tired of the game and want your help. Given N, create a valid sequence of states that solves the game. If there are multiple solutions return any one.

PROBLEM NAME: rocks

INPUT FORMAT:
* Line 1: A single integer: N

SAMPLE INPUT (file rocks.in):
3

OUTPUT FORMAT:
* Lines 1..2^N+1: A string of length N containing only 'O' and 'X' (where O denotes a uncovered hole and X denotes a covered hole). The jth character in this line represents whether the jth hole is covered or uncovered in this state. The first and last lines must be all uncovered (all O).

SAMPLE OUTPUT (file rocks.out):
OOO
OXO
OXX
OOX
XOX
XXX
XXO
XOO
OOO

题目大意就是写出长度为n的X和O的所有排列,其中相邻的两个排列之间只能有一个数不同。

因为数据不是很大最多有2 ^ 15种排列,所以dfs就行。

只要每一次改变其中一个数,然后判断这种排列前面是否已经存在过,不存在就输出。

但每一次判断的时候是一个字符串,无论是空间上还是时间上都不是很好。于是有一个优化:吧‘O’看成0,'X'看成1,于是就变成了一个01串,即一个数的二进制。

于是我们只要尝试修改这个数的每一位,然后判断得到的新的数是否存在过就行。

至于修改,当然是亦或号啦

 #include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cctype> //isdigit
using namespace std;
typedef long long ll;
#define enter printf("\n")
const int maxn = 1e6 + ;
const int INF = 0x3f3f3f3f;
inline ll read()
{
ll ans = ;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) {last = ch; ch = getchar();}
while(isdigit(ch))
{
ans = ans * + ch - ''; ch = getchar();
}
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < ) {putchar('-'); x = -x;}
if(x == ) {putchar(''); return;}
int q[], N = ;
q[] = ;
while(x) {q[++N] = x % ; x /= ;}
while(N) {putchar('' + q[N]); --N;}
} int n;
bool vis[maxn];
void print(int x) //从高位开始输出每一位
{
vis[x] = ;
for(int i = n - ; i >= ; --i)
{
if((x >> i) & ) printf("X");
else printf("O");
}
enter;
}
void dfs(int step, int x)
{
if(!vis[x]) print(x); //之所以放在这,而不是第53行之后,是为了输出最开始的OOOOOOO情况
if(step == ( << n) + ) exit();
for(int i = ; i <= n - ; ++i)
{
int now = x ^ ( << i);
if(!vis[now]) dfs(step + , now);
}
} int main()
{
n = read();
dfs(, );
for(int i = ; i <= n; ++i) printf("O"); enter;
return ;
}

The Rock Game的更多相关文章

  1. ural 2069. Hard Rock

    2069. Hard Rock Time limit: 1.0 secondMemory limit: 64 MB Ilya is a frontman of the most famous rock ...

  2. POJ - 2339 Rock, Scissors, Paper

    初看题目时就发了个错误,我因为没有耐心看题而不了解题目本身的意思,找不到做题的突破口,即使看了一些题解,还是没有想到方法. 后来在去问安叔,安叔一语道破天机,问我有没有搞清题目的意思,我才恍然大悟,做 ...

  3. ROCK 聚类算法‏

    ROCK (RObust Clustering using linKs)  聚类算法‏是一种鲁棒的用于分类属性的聚类算法.该算法属于凝聚型的层次聚类算法.之所以鲁棒是因为在确认两对象(样本点/簇)之间 ...

  4. Rice Rock

    先翻译评分要点,然后一点点翻译程序实现过程 如何产生一堆岩石? rock_group = set([])#空集合,全局变量   rock_group.add(a_rock) 要画出来draw hand ...

  5. HDOJ(HDU) 2164 Rock, Paper, or Scissors?

    Problem Description Rock, Paper, Scissors is a two player game, where each player simultaneously cho ...

  6. Hard Rock

    Ilya is a frontman of the most famous rock band on Earth. Band decided to make the most awesome musi ...

  7. 弹指之间 -- Folk Rock

    CHAPTER 17 民谣摇滚 Folk Rock 以8Beat为主,120左右的速度最能表现此节奏特色. 示例曲目: 略

  8. 2018 ACM-ICPC 中国大学生程序设计竞赛线上赛 H题 Rock Paper Scissors Lizard Spock.(FFT字符串匹配)

    2018 ACM-ICPC 中国大学生程序设计竞赛线上赛:https://www.jisuanke.com/contest/1227 题目链接:https://nanti.jisuanke.com/t ...

  9. HDU 2164 Rock, Paper, or Scissors?

    http://acm.hdu.edu.cn/showproblem.php?pid=2164 Problem Description Rock, Paper, Scissors is a two pl ...

随机推荐

  1. 关于sublimeText3 设置格式化代码快捷键的问题

    sublime中自建的有格式化按钮: Edit  ->  Line  ->  Reindent 只是sublime并没有给他赋予快捷键,所以只需加上快捷键即可 Preference  -& ...

  2. extJs常用的四种Ajax异步提交

    /** * 第一种Ajax提交方式 * 这种方式需要直接使用ext Ajax方法进行提交 * 使用这种方式,需要将待传递的参数进行封装 * @return */function saveUser_aj ...

  3. JUC源码1-原子量

    什么是原子量,原子量就是一次操作,要么成功,要么失败.比如java中的i++ 或i-- , 不具备原子性,每次读取的值都是不一样的,探究其原因,x86体系中,他的总线是32位的,i++的操作指令必须是 ...

  4. 【NOI2000】 单词查找树

    问题描述 在进行文法分析的时候,通常需要检测一个单词是否在我们的单词列表里.为了提高查找和定位的速度,通常都画出与单词列表所对应的单词查找树,其特点如下: 根结点不包含字母,除根结点外每一个结点都仅包 ...

  5. ubuntu16.04安装ssh服务,并实现远程访问

    一.查看是否安装了ssh服务 apt-cache policy openssh-client openssh-server ubuntu默认安装了openssh-client,openssh-serv ...

  6. LeetCode DB: Department Highest Salary

    The Employee table holds all employees. Every employee has an Id, a salary, and there is also a colu ...

  7. python-适配器模式

    源码地址:https://github.com/weilanhanf/PythonDesignPatterns 说明: 为了解决接口不兼容的问题引进一种接口的兼容机制,就是适配器模式,其通过提供一种适 ...

  8. python乐观锁、悲观锁

    二.乐观锁总是认为不会产生并发问题,每次去取数据的时候总认为不会有其他线程对数据进行修改,因此不会上锁,但是在更新时会判断其他线程在这之前有没有对数据进行修改 三.悲观锁总是假设最坏的情况,每次取数据 ...

  9. 关于对DI和IOC的概念理解

    在spring框架学习过程中,涉及到两个新名词:DI和IOC.开始总是混淆两者的概念,稀里糊涂,后来上网搜了一下又和同学讨论之后,基本上理解了二者的概念.实际上DI(依赖注入)和IOC(控制反转)就是 ...

  10. Nginx 优化配置及详细注释

    Nginx 的nginx.conf文件,是调优后的,具体影响已经写清楚注释,可以拿来用,有一些设置无效,我备注上了,不知道是不是版本的问题,回头查一下再更正. #普通配置 #==性能配置 #运行用户 ...