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
思路:dfs
实现代码:
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<set>
#include<list>
using namespace std;
#define ll long long
#define sd(x) scanf("%d",&x)
#define sdd(x,y) scanf("%d%d",&x,&y)
#define sddd(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define sf(x) scanf("%s",x)
#define ff(i,x,y) for(int i = x;i <= y;i ++)
#define fj(i,x,y) for(int i = x;i >= y;i --)
#define mem(s,x) memset(s,x,sizeof(s));
#define pr(x) printf("%d",x);
const int Mod = 1e9+;
const int inf = 1e9;
const int Max = 1e5+;
void exgcd(ll a,ll b,ll& d,ll& x,ll& y){if(!b){d=a;x=;y=;}else{exgcd(b,a%b,d,y,x);y-=x*(a/b);}}
ll inv(ll a,ll n){ll d, x, y;exgcd(a,n,d,x,y);return (x+n)%n;}
int gcd(int a,int b) { return (b>)?gcd(b,a%b):a; }
int lcm(int a, int b) { return a*b/gcd(a, b); }
//int mod(int x,int y) {return x-x/y*y;}
char s[];
int mp[][];
int ans = inf;
int check()
{
int x = mp[][];
for(int i = ;i < ;i ++){
for(int j = ;j < ;j ++){
if(mp[i][j]!=x)
return ;
}
}
return ;
}
void fun(int x,int y)
{
mp[x][y] = !mp[x][y];
if(x - >= ) mp[x-][y] = !mp[x-][y];
if(x + <= ) mp[x+][y] = !mp[x+][y];
if(y - >= ) mp[x][y-] = !mp[x][y-];
if(y + <= ) mp[x][y+] = !mp[x][y+];
}
int dfs(int x,int y,int t)
{
if(check()){
ans = min(ans,t);
return ;
}
if(x>=||y>=)
return ;
int fx = (x+)%;
int fy = y+(x+)/;
dfs(fx,fy,t);
//cout<<fx<<" "<<fy<<" "<<t<<endl;
fun(x,y); dfs(fx,fy,t+);
//cout<<fx<<" "<<fy<<" "<<t+1<<endl;
fun(x,y);
return ;
}
int main()
{
for(int i=;i<;i++){
cin>>s;
for(int j=;j<;j++){
if(s[j]=='b')
mp[i][j] = ;
else
mp[i][j] = ;
}
}
dfs(,,);
if(ans==inf)
cout<<"Impossible"<<endl;
else
cout<<ans<<endl;
return ;
}

poj1753 【枚举】的更多相关文章

  1. poj1753枚举

    Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 33670   Accepted: 14713 Descr ...

  2. POJ-1753 Flip Game---二进制枚举子集

    题目链接: https://vjudge.net/problem/POJ-1753 题目大意: 有4*4的正方形,每个格子要么是黑色,要么是白色,当把一个格子的颜色改变(黑->白或者白-> ...

  3. POJ1753 Flip Game(bfs、枚举)

    链接:http://poj.org/problem?id=1753 Flip Game Description Flip game is played on a rectangular 4x4 fie ...

  4. [POJ1753]Flip Game(异或方程组,高斯消元,枚举自由变量)

    题目链接:http://poj.org/problem?id=1753 题意:同上. 这回翻来翻去要考虑自由变元了,假设返回了自由变元数量,则需要枚举自由变元. /* ━━━━━┒ギリギリ♂ eye! ...

  5. poj1753解题报告(枚举、组合数)

    POJ 1753,题目链接http://poj.org/problem?id=1753 题意: 有4*4的正方形,每个格子要么是黑色,要么是白色,当把一个格子的颜色改变(黑->白或者白-> ...

  6. poj-3279 poj-1753(二进制枚举)

    题目链接:http://poj.org/problem?id=3279 题目大意: 有一个m*n的棋盘(1 ≤ M ≤ 15; 1 ≤ N ≤ 15),每个格子有两面分别是0或1,每次可以对一个格子做 ...

  7. 二进制枚举例题|poj1222,poj3279,poj1753

    poj1222,poj3279,poj1753 听说还有 POJ1681-画家问题 POJ1166-拨钟问题 POJ1054-讨厌的青蛙

  8. POJ1753 Flip Game(位运算+暴力枚举)

    Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 square ...

  9. POJ1753(位操作和枚举)

    题目:http://poj.org/problem?id=1753 题意:一块4*4的棋盘,黑白块不规律分布,翻动一个色块,其上下左右,都会被翻动,知道全黑全白为止.输出最小次数,达不到则输出“Imp ...

  10. [POJ1753]Flip Game(开关问题,枚举)

    题目链接:http://poj.org/problem?id=1753 和上一个题一样,将初始状态存成01矩阵,就可以用位运算优化了.黑色白色各来一遍 /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏ ...

随机推荐

  1. python wsgi 简介

    wsgi全称是"Web Server Gateway Interfacfe",web服务器网关接口,wsgi在python2.5中加入,是web服务器和web应用的标准接口,任何实 ...

  2. 一篇文章让你彻底掌握 shell 语言

    一篇文章让你彻底掌握 shell 语言 由于 bash 是 Linux 标准默认的 shell 解释器,可以说 bash 是 shell 编程的基础. 本文主要介绍 bash 的语法,对于 linux ...

  3. Linux shell ftp命令下载文件 根据文件日期

    需求:ftp获取远程数据的文件,根据文件的创建时间点下载文件. 可以自行扩展根据文件的大小等其他需求. 知识点总结: 1.获取文件的时间: ls -lrt|awk '{print $6" & ...

  4. Luogu3676 小清新数据结构题 动态点分治

    传送门 换根类型的统计问题动态点分治都是很好做的. 设所有点的点权和为$sum$ 首先,我们先不考虑求$\sum\limits_i s_i^2$,先考虑如何在换根的情况下求$\sum\limits_i ...

  5. HNOI2018简要题解

    HNOI2018简要题解 D1T1 寻宝游戏 题意 某大学每年都会有一次 Mystery Hunt 的活动,玩家需要根据设置的线索解谜,找到宝藏的位置,前一年获胜的队伍可以获得这一年出题的机会. 作为 ...

  6. 限流——spring-cloud-zuul-ratelimit

    先留个坑,慢慢补 git代码Demo:https://github.com/islowcity/spring-cloud-zuul-ratelimiter.git 有时间再写分析

  7. 【JVM.11】Java内存模型与线程

    鲁迅曾经说过“并发处理的广泛应用是使得Amdahl定律代替摩尔定律成为计算机性能发展源动力的根本原因,也是人类‘压榨‘ 计算机运行能力的最有力武器.” 一.概述 多任务处理在现代计算机操作系统中几乎已 ...

  8. 仓储层接口IBaseRepository解析

    //代码调用由业务层调用,调用方式详见源代码的业务层,升级直接替换TT模板即可,无需覆盖系统using System; using System.Collections.Generic; using ...

  9. Session配置之WebApi支持

    1.在WebApiConfig中建立建立HttpControllerHandler和HttpControllerRouteHandler 并覆写它 public class SessionRouteH ...

  10. db2安装

    官网下载: DB2 11.1 data server trial for Linux® on AMD64 and Intel® EM64T systems (x64)v11.1_linuxx64_se ...