(枚举,dfs)The Pilots Brothers' refrigerator poj 2965
来源:互联网 发布:亚洲人讲英语 知乎 编辑:程序博客网 时间:2024/06/03 14:40
The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.
There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location[i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in rowi and all handles in column j.
The task is to determine the minimum number of handle switching necessary to open the refrigerator.
The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.
The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.
-+-----------+--
61 11 31 44 14 34 4
#include<iostream>#include<cstdio>#include<cstring>using namespace std;char s[15][15];int a[15][15];int alt[15][15];int num;int ans[15][15];int judge(){ for(int i=1; i<=4; ++i) for(int j=1; j<=4; ++j) { if(!a[i][j]) return 0; } return 1;}void fil(int x,int y){ alt[x][y]=!alt[x][y]; for(int i=1; i<=4; ++i) a[x][i]=!a[x][i]; for(int i=1; i<=4; ++i) a[i][y]=!a[i][y]; a[x][y]=!a[x][y];}void copyans(){ for(int i=1; i<=4; ++i) for(int j=1; j<=4; ++j) ans[i][j]=alt[i][j];}int dfs(int x,int y,int cnt){ if(judge()) { num=min(num,cnt); copyans(); return 0; } if(x>4) return 0; int ny=(y+1)%5; int nx=x+(y+1)/5; if(ny==0) ny=1; dfs(nx,ny,cnt); fil(x,y); dfs(nx,ny,cnt+1); fil(x,y); return 0;}void put(){ for(int i=1; i<=4; ++i) for(int j=1; j<=4; ++j) if(ans[i][j]) cout<<i<<" "<<j<<endl;}int main(){ for(int i=1; i<=4; ++i) scanf("%s",s[i]+1); for(int i=1; i<=4; ++i) for(int j=1; j<=4; ++j) { if(s[i][j]=='+') a[i][j]=0; else a[i][j]=1; } memset(alt,0,sizeof(alt)); num=999999; dfs(1,1,0); cout<<num<<endl; put();}
- POJ 2965 The Pilots Brothers' refrigerator(枚举+dfs)
- POJ 2965 The Pilots Brothers' refrigerator(枚举+DFS)
- POJ 2965 The Pilots Brothers' refrigerator (DFS + 枚举)
- (枚举,dfs)The Pilots Brothers' refrigerator poj 2965
- POJ 2965 The Pilots Brothers' refrigerator【枚举+dfs】
- POJ 2965 The Pilots Brothers' refrigerator(dfs+枚举 || 规律)
- POJ 2965 The Pilots Brothers' refrigerator 枚举dfs
- POJ 2965-The Pilots Brothers' refrigerator(枚举&&DFS&&输出过程)
- The Pilots Brothers' refrigerator(POJ 2965)(dfs枚举+状态压缩)
- POJ 2965 The Pilots Brothers' refrigerator(枚举+dfs)
- POJ 2965 The Pilots Brothers' refrigerator 枚举
- POJ 2965 The Pilots Brothers' refrigerator(枚举)
- POJ 2965 The Pilots Brothers' refrigerator 枚举
- poj 2965 The Pilots Brothers' refrigerator (枚举)
- poj 2965 The Pilots Brothers' refrigerator[ 枚举 ]
- poj-2965 The Pilots Brothers' refrigerator -- 枚举
- poj 2965 The Pilots Brothers' refrigerator【枚举】
- The Pilots Brothers' refrigerator 枚举 Poj 2965
- java基础知识(五)
- 测试图片
- Strlen与Strcpy的模拟实现
- channel代码示例
- linux下安装pygame错误:linux/videodev.h:No such file or directory error解决方法
- (枚举,dfs)The Pilots Brothers' refrigerator poj 2965
- 【mark,计划2017年底开始用Linux~】完全用 Gnu/Linux 工作
- Centos7 配置静态IP
- 【C++学习笔记】对float型数据类型和double型数据的理解
- Linux下安装mysql数据库
- openGL学习(二)
- 消息通知之Notification与PendingIntent<一>
- Codeforces 469C 24 Game【思维+模拟】
- Android中实现波浪球效果