Flip Game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 28805   Accepted: 12461

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

 //164K    125MS    C++    1211B    2014-04-26 11:02:12
/* 题意:
问最少翻几步可以使棋盘棋子一样,不可能就输出Impossible 搜索枚举:
枚举全部状态,每个位置的棋子有翻或不翻两种状态,枚举全部状态。
注意一个棋子翻两次则和没翻一样,所以一种有2^16种情况,用dfs枚举全部状态。 */
#include<stdio.h>
#include<string.h>
int g[][];
int flag;
int judge(int tg[][]) //判断
{
for(int i=;i<=;i++)
for(int j=;j<=;j++)
if(g[i][j]!=g[][]) return ;
return ;
}
void flip(int i,int j) //翻棋
{
g[i][j]^=;
g[i-][j]^=;
g[i+][j]^=;
g[i][j-]^=;
g[i][j+]^=;
}
void dfs(int x,int y,int cnt,int n)
{
if(cnt==n){
flag=judge(g);
return;
}
if(flag || y>) return;
flip(x,y);
if(x<) dfs(x+,y,cnt+,n);
else dfs(,y+,cnt+,n);
flip(x,y);
if(x<) dfs(x+,y,cnt,n);
else dfs(,y+,cnt,n);
}
int main(void)
{
char c[];
while(scanf("%s",c)!=EOF)
{
memset(g,,sizeof(g));
for(int i=;i<;i++) g[][i+]=c[i]=='b'?:;
for(int i=;i<;i++){
scanf("%s",c);
for(int j=;j<;j++)
g[i+][j+]=c[j]=='b'?:;
}
flag=;
int cnt=-;
for(int i=;i<;i++){
dfs(,,,i);
if(flag){
cnt=i;break;
}
}
if(cnt==-) puts("Impossible");
else printf("%d\n",cnt);
}
return ;
}
/* bwwb
bbwb
bwwb
bwww bwbw
bwww
wwwb
wwwb bwww
wwww
wwww
wwww */

poj 1753 Flip Game (dfs)的更多相关文章

  1. POJ 1753 Flip Game DFS枚举

    看题传送门:http://poj.org/problem?id=1753 DFS枚举的应用. 基本上是参考大神的.... 学习学习.. #include<cstdio> #include& ...

  2. POJ 1753 Flip Game (DFS + 枚举)

    题目:http://poj.org/problem?id=1753 这个题在開始接触的训练计划的时候做过,当时用的是DFS遍历,其机制就是把每一个棋子翻一遍.然后顺利的过了.所以也就没有深究. 省赛前 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. poj 1753 Flip Game(bfs状态压缩 或 dfs枚举)

    Description Flip game squares. One side of each piece is white and the other one is black and each p ...

  7. OpenJudge/Poj 1753 Flip Game

    1.链接地址: http://bailian.openjudge.cn/practice/1753/ http://poj.org/problem?id=1753 2.题目: 总时间限制: 1000m ...

  8. POJ 1753 Flip Game(状态压缩+BFS)

    题目网址:http://poj.org/problem?id=1753 题目: Flip Game Description Flip game is played on a rectangular 4 ...

  9. poj 1753 Flip Game 枚举(bfs+状态压缩)

    题目:http://poj.org/problem?id=1753 因为粗心错了好多次……,尤其是把1<<15当成了65535: 参考博客:http://www.cnblogs.com/k ...

随机推荐

  1. 汇编:采用址表的方法编写程序实现C程序的switch功能

    //待实现的C程序 1 void main() { ; -) { : printf("excellence"); break; : printf("good") ...

  2. jQuery 使用问题

    attr('checked', 'checked')调用多次仅第一次生效 使用attr()获取这些属性的返回值为String类型,如果被选中(或禁用)就返回checked.selected或disab ...

  3. php-5.6.26源代码 - hash存储结构 - 添加

    添加 , (void *)module, sizeof(zend_module_entry), (void**)&module_ptr){ // zend_hash_add 定义在文件“php ...

  4. tp5 数据库信息导出到excel(带图片)

    function excel_down(){ //导入谁就去查谁 $data=Db::name('order_xueyou')->select(); // 导出Exl // import(&qu ...

  5. docker使用命令汇总

    docker命令 docker ps 容器列表 docker ps -a 所有容器列表,包含未运行的容器 docker image ls 镜像列表 docker logs -f xxx 容器日志 do ...

  6. Windows10 快捷键

    windows 10快捷键: F1 打开帮助 F2 重命名 F3 打开搜索文件和文件夹 F4 打开地址栏常用的地址 F5 刷新 F11   全屏 选择文件和内容: shift + 上下左右键选择连续的 ...

  7. poj 2393 奶牛场生产成本问题 贪心算法

    题意:有一个奶牛场,第i周的生产成本为c,需要数量为 y,每周的存储成本为s.问怎么安排使得成本最低? 思路: 成本最低是吧?求出每周的最低成本*该周需要的数量就是成本最低 每周的成本有两个:自己本周 ...

  8. C++各种类型的简单排序大汇总~

    啊,排序的技能点也太多了吧!!!LITTLESUN快要**在排序的技能场了啊!(划掉)经历了两天48小时2880分钟172800秒的艰苦奋斗,终于终于终于学的差不多了!明天就可以去打排序的小怪喽!(撒 ...

  9. spring boot 入门3 如何在springboot 上使用AOP

    Aop是spring的两大核心之一 那么如何在springboot中采用注解的形式实现aop那? 1)首先我们定义一个相关功能的切面类 并 采用@Aspect 注解来声明当前类为切面 同时采用@Com ...

  10. itop-4412开发板使用第一篇-信号量的学习使用

    1. 本次基于itop-4412研究下Linux信号量的使用方法. 2. 创建信号量的函数,信号量的头文件在那个路径?编译应用程序的话,头文件有3个路径,内核源码头文件,交叉编译器头文件,ubuntu ...