Flip Game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 59468   Accepted: 24750

Description

Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules: 
  1. Choose any one of the 16 pieces.
  2. Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).

Consider the following position as an example:

bwbw 
wwww 
bbwb 
bwwb 
Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become:

bwbw 
bwww 
wwwb 
wwwb 
The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal. 

Input

The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.

Output

Write to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it's impossible to achieve the goal, then write the word "Impossible" (without quotes).

Sample Input

bwwb
bbwb
bwwb
bwww

Sample Output

4

Source

 
 
 #include <iostream>
#include <cstdio> using namespace std; char str[];
int a[][];
int ans=; void flip(int x,int y)
{
a[x][y]=!a[x][y];
a[x-][y]=!a[x-][y];
a[x+][y]=!a[x+][y];
a[x][y-]=!a[x][y-];
a[x][y+]=!a[x][y+];
} bool same_color(void)
{
for(int i=;i<=;++i)
{
for(int j=;j<=;++j)
{
if(a[i][j]!=a[][])
{
return false;
}
}
}
return true;
} void dfs(int x,int y,int t)
{
// 判断是否同色,同色满足,返回
if(same_color())
{
if(ans>t)
{
ans=t;
}
return;
}
// 深搜截止条件
if(x>)
{
return;
} if(y==)
{
// 当前不反转
dfs(x+,,t);
// 当前反转
flip(x,y);
dfs(x+,,t+);
// 状态还原
flip(x,y);
}
else
{
dfs(x,y+,t);
flip(x,y);
dfs(x,y+,t+);
flip(x,y);
} } int main()
{
for(int i=;i<=;++i)
{
scanf("%s",str+);
for(int j=;j<=;++j)
{
// a->0,b->1
if(str[j]=='w')
{
a[i][j]=;
}
else
{
a[i][j]=;
}
//a[i][j]=str[j]-'a';
//printf("%d ",a[i][j]);
}
//printf("\n");
} dfs(,,); if(ans==)
{
printf("Impossible\n");
}
else
{
printf("%d\n",ans);
} return ;
}

POJ 1753 Flip Game 暴力 深搜的更多相关文章

  1. poj 1753 Flip Game(暴力枚举)

    Flip Game   Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 52279   Accepted: 22018 Des ...

  2. 枚举 POJ 1753 Flip Game

    题目地址:http://poj.org/problem?id=1753 /* 这题几乎和POJ 2965一样,DFS函数都不用修改 只要修改一下change规则... 注意:是否初始已经ok了要先判断 ...

  3. POJ 1753. Flip Game 枚举or爆搜+位压缩,或者高斯消元法

    Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37427   Accepted: 16288 Descr ...

  4. POJ - 1753 Flip Game(状压枚举)

    https://vjudge.net/problem/POJ-1753 题意 4*4的棋盘,翻转其中的一个棋子,会带动邻接的棋子一起动.现要求把所有棋子都翻成同一种颜色,问最少需要几步. 分析 同一个 ...

  5. Poj(2488),按照字典序深搜

    题目链接:http://poj.org/problem?id=2488 思路:按照一定的字典序深搜,当时我的想法是把所有的可行的路径都找出来,然后字典序排序. 后来,凡哥说可以在搜索路径的时候就按照字 ...

  6. POJ 1753 Flip Game(高斯消元+状压枚举)

    Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 45691   Accepted: 19590 Descr ...

  7. HDU--杭电--1195--Open the Lock--深搜--都用双向广搜,弱爆了,看题了没?语文没过关吧?暴力深搜难道我会害羞?

    这个题我看了,都是推荐的神马双向广搜,难道这个深搜你们都木有发现?还是特意留个机会给我装逼? Open the Lock Time Limit: 2000/1000 MS (Java/Others)  ...

  8. POJ 1753 Flip Game (状压+暴力)

    题目链接:http://poj.org/problem?id=1753 题意: 给你一个4*4的棋盘,上面有两种颜色的棋子(一种黑色,一种白色),你一次可以选择一个棋子翻转它(黑色变成白色,同理反之) ...

  9. POJ 1753 Flip Game 状态压缩,暴力 难度:1

    Flip Game Time Limit: 1000MS  Memory Limit: 65536K  Total Submissions: 4863  Accepted: 1983 Descript ...

随机推荐

  1. 代码审计之CVE-2017-6920 Drupal远程代码执行漏洞学习

     1.背景介绍: CVE-2017-6920是Drupal Core的YAML解析器处理不当所导致的一个远程代码执行漏洞,影响8.x的Drupal Core. Drupal介绍:Drupal 是一个由 ...

  2. JAVA中值传递,引用传递

    刚在写一个用例,需要在方法中改变传递的参数的值,可是java中只有传值调用,没有传址调用.所以在java方法中改变参数的值是行不通的.但是可以改变引用变量的属性值. 可以仔细理解一下下面几句话: 1. ...

  3. python接口自动化测试 - openpyxl封装类

    前言 为了更好的让openpyxl在工作中使用,将openpyxl的常用操作封装起来,这样不仅复用性高,而且阅读性好 直接上代码 #!/usr/bin/env python # -*- coding: ...

  4. 团队项目—Beta版本冲刺(2/3)

    团队信息 何全江(队长) 201731024218 胡志伟 201731024240 李元港 201731024232 孟诚成 201731024242 罗俊杰 201731024226 雷安勇 20 ...

  5. 创建dynamics CRM client-side (八) - 获取attribute的值 和 设置disable

    大家可以用下面的方式来获取attribute的值 formContext.getAttribute("address1_shippingmethodcode").getText() ...

  6. jenkins 集成jmeter-简单篇

    测试用例上传至gitlab后,使用jenkins集成gitlab,并执行压测命令 执行完成后,可在jenkins中查看压测报告不同的项目创建最好创建不同的project) [集成]安装&配置& ...

  7. iperf安装使用教程

    https://linoxide.com/monitoring-2/install-iperf-test-network-speed-bandwidth/

  8. linux--->阿里云centos6.9环境配置安装lnmp

    阿里云centos6.9环境配置安装lnmp mysql安装 本人博客:http://www.cnblogs.com/frankltf/p/8615418.html PHP安装 1.安装依赖关系 yu ...

  9. AcWing 789.数的范围

    AcWing 789.数的范围 题目描述 给定一个按照升序排列的长度为n的整数数组,以及 q 个查询. 对于每个查询,返回一个元素k的起始位置和终止位置(位置从0开始计数). 如果数组中不存在该元素, ...

  10. 我的一个react路由之旅(步骤及详图)

    今天开始react一个重要部分的xiao~习,路由~(过程截图,最后附代码) 以下代码只能骗糊涂蛋子,没错,就是我自己,不要打算让我敲出多高级的东西~ 理论性知识几乎没有,请不要打算让我给你说原理啥的 ...