Flip Game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 40632   Accepted: 17647

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

题目地址:http://poj.org/problem?id=1753

#include<stdio.h>
#include<string.h>
#include<iostream> #include<stdlib.h> using namespace std; #define N 7 bool mat[N][N], flag;
int deep; int dx[] = {, -, , , };
int dy[] = {, , , -, }; void Init()
{
char c;
for(int i = ; i <= ; i++)
{
for(int j = ; j <= ; j++)
{
scanf(" %c",&c);
if(c == 'b')
{
mat[i][j] = ;
}
else
{
mat[i][j] = ;
}
}
}
}
void Print()
{
for(int i = ; i <= ; i++)
{
for(int j = ; j <= ; j++)
printf("%d", mat[i][j]);
printf("\n");
}
} void Change(int x, int y)
{
int next_x, next_y;
for(int k = ; k < ; k++)
{
next_x = x + dx[k];
next_y = y + dy[k];
//if(next_x >= 1 && next_x <=4 && next_y >= 1 && next_y <=4)
//{
mat[next_x][next_y] = !mat[next_x][next_y];
//}
}
}
void Flip(int x)
{
switch(x) //1~16种case代表4*4的16个格子
{ case : Change(,); break;
case : Change(,); break;
case : Change(,); break;
case : Change(,); break; case : Change(,); break;
case : Change(,); break;
case : Change(,); break;
case : Change(,); break; case : Change(,); break;
case : Change(,); break;
case : Change(,); break;
case : Change(,); break; case : Change(,); break;
case : Change(,); break;
case : Change(,); break;
case : Change(,); break;
}
}
bool Result()
{
for(int i = ; i <= ; i++)
{
for(int j = ; j <= ; j++)
if(mat[i][j] != mat[][])
return false;
}
return true;
}
void Dfs(int x, int dp)
{ if(flag || dp > deep || x>) return;
//printf("DFS(%d, %d)\n", x, dp); Flip(x);
//Print();printf("---------\n");
if(dp == deep)
{
flag = (flag || Result());
if(flag)
{
//printf("OK!!!!!!!!!!!!!!");
//exit(0);
goto A;
} }
Dfs(x+, dp+);
Flip(x);
//Print();printf("---------\n");
Dfs(x+, dp);
A: return;
} int main()
{
//每个棋子最多翻一次(其实是奇数次,但是没意义),翻偶数次和没翻一样,所以每个棋子就两种状态,翻或者不翻
//所以一共就有2^16次方种可能,枚举这些可能就行了,DFS
//从0到16,如果16还找不到那就是Impossible
int a, b;
Init();
//Print(); flag = false;
if(Result())
{
cout<<<<endl;
return ;
} for(deep = ; deep <= ; deep++)
{
//cout<<"deep = "<<deep<<endl;
Dfs(, );
if(flag) break;
} if(flag) cout<<deep<<endl;
else cout<<"Impossible"<<endl; // while(cin>>a)
// {
// Flip(a);
// Print();
// } return ;
}
/*
bwwb
bbwb
bwwb
bwww wwww
wwww
wwww
wwww bbbw
wbww
wwww
wwww wwww
wwww
wwwb
wwbb bbww
bwww
wwww
wwww wwbw
bbww
wwww
wwww
*/

POJ 1753 (枚举+DFS)的更多相关文章

  1. POJ 3050 枚举+dfs+set判重

    思路: 枚举+搜一下+判个重 ==AC //By SiriusRen #include <set> #include <cstdio> using namespace std; ...

  2. 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 ...

  3. [ACM训练] 算法初级 之 基本算法 之 枚举(POJ 1753+2965)

    先列出题目: 1.POJ 1753 POJ 1753  Flip Game:http://poj.org/problem?id=1753 Sample Input bwwb bbwb bwwb bww ...

  4. 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 ...

  5. POJ 2965 The Pilots Brothers' refrigerator【枚举+dfs】

    题目:http://poj.org/problem?id=2965 来源:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=26732#pro ...

  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. poj 1753 2965

    这两道题类似,前者翻转上下左右相邻的棋子,使得棋子同为黑或者同为白.后者翻转同行同列的所有开关,使得开关全被打开. poj 1753 题意:有一4x4棋盘,上面有16枚双面棋子(一面为黑,一面为白), ...

  8. POJ 1222 POJ 1830 POJ 1681 POJ 1753 POJ 3185 高斯消元求解一类开关问题

    http://poj.org/problem?id=1222 http://poj.org/problem?id=1830 http://poj.org/problem?id=1681 http:// ...

  9. POJ.3172 Scales (DFS)

    POJ.3172 Scales (DFS) 题意分析 一开始没看数据范围,上来直接01背包写的.RE后看数据范围吓死了.然后写了个2^1000的DFS,妥妥的T. 后来想到了预处理前缀和的方法.细节以 ...

随机推荐

  1. Android中动态改变控件的大小的一种方法

    在Android中有时候我们需要动态改变控件的大小.有几种办法可以实现  一是在onMeasure中修改尺寸,二是在onLayout中修改位置和尺寸.这个是可以进行位置修改的,onMeasure不行. ...

  2. read(byte[] b)与readFully(byte[] b)

    转载于:http://yyzjava.iteye.com/blog/1178525 要搞清楚read(byte[] b)和readFully(byte[] b)的区别,可以从以下方面着手分析: 1.代 ...

  3. [APIO2009]抢掠计划 ($Tarjan$,最长路)

    题目链接 Solution 裸题诶... 直接 \(Tarjan\) 缩点+ \(SPFA\) 最长路即可. 不过在洛谷上莫名被卡... RE两个点... Code #include<bits/ ...

  4. hdu 3625 Examining the Rooms 轮换斯特林数

    题目大意 n个房间对应n把钥匙 每个房间的钥匙随机放在某个房间内,概率相同. 有K次炸门的机会,求能进入所有房间的概率 一号门不给你炸 分析 我们设\(key_i\)为第i间房里的钥匙是哪把 视作房间 ...

  5. 2018.8.7 Noip2018模拟测试赛(二十)

    日期: 八月七号  总分: 300分  难度: 提高 ~ 省选    得分: 100分(呵呵一笑) 题目列表: T1:SS T2:Tree Game T3:二元运算 赛后反思: Emmmmmm…… 开 ...

  6. sublime flatLand 主题

    今天试了下感觉主题不错 记下来备忘. 1.sublime3 package control install  搜索 flatLand 2 安装完成后. 修改 Preferences 文件,通过 Sub ...

  7. 五、 java中数组

    定义数组的两种方式 class myarray1 { public static void main(String[] args) { //1.如何定义一个数组 //1.1数组的声明 String[] ...

  8. hdu 1180(广搜好题)

    诡异的楼梯 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Subm ...

  9. 使用Naive Bayes从个人广告中获取区域倾向

    RSS源介绍:https://zhidao.baidu.com/question/2051890587299176627.html http://www.rssboard.org/rss-profil ...

  10. pycharm上传代码到码云(详细)

    如要转载 麻烦请您备注好原文出处!!!!(谢谢合作!) >>首先要去码云注册个账号 提示(尽量使用英文名)创建用户名 使用邮箱登录 >>然后创建库  >填写项目的基础信息 ...