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 row i and all handles in column j.

The task is to determine the minimum number of handle switching necessary to open the refrigerator.

Input

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.

Output

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.

Sample Input

-+--
----
----
-+--

Sample Output

6
1 1
1 3
1 4
4 1
4 3
4 4
思路:dfs,加个数组保存路径
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,i,j;
int rankx[],ranky[],fx[],fy[];
int check()
{
for(i=;i<;i++){
for(j=;j<;j++){
if(mp[i][j]!=)
return ;
}
}
return ;
} void fan(int x,int y){
for(i=;i<;i++){
mp[x][i] = !mp[x][i];
mp[i][y] = !mp[i][y];
}
mp[x][y] = !mp[x][y];
} int dfs(int x,int y,int t)
{
if(check()){
if(ans>t){
ans = t;
for(i=;i<t;i++){
rankx[i] = fx[i];
ranky[i] = fy[i];
}
}
return ;
}
if(x>=||y>=)
return ;
int nx = (x+)%;
int ny = y+(x+)/;
dfs(nx,ny,t);
fan(x,y); fx[t] = x+;
fy[t] = y+; dfs(nx,ny,t+);
fan(x,y); return ;
}
int main()
{
for(i=;i<;i++){
cin>>s;
for(j=;j<;j++){
if(s[j]=='+')
mp[i][j] = ;
else
mp[i][j] = ;
}
}
dfs(,,);
cout<<ans<<endl;
for(i=;i<ans;i++){
cout<<rankx[i]<<" "<<ranky[i]<<endl;
}
return ;
}

poj2965 【枚举】的更多相关文章

  1. poj2965枚举

    The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20398 ...

  2. 假期训练八(poj-2965递归+枚举,hdu-2149,poj-2368巴什博奕)

    题目一(poj-2965):传送门 思路:递归+枚举,遍历每一种情况,然后找出最小步骤的结果,与poj-1753类似. #include<iostream> #include<cst ...

  3. poj2965 The Pilots Brothers&#39; refrigerator(直接计算或枚举Enum+dfs)

    转载请注明出处:http://blog.csdn.net/u012860063? viewmode=contents 题目链接:http://poj.org/problem? id=2965 ---- ...

  4. dfs+枚举,flip游戏的拓展POJ2965

    POJ 2965             The Pilots Brothers' refrigerator Description The game “The Pilots Brothers: fo ...

  5. POJ-2965 The Pilots Brothers' refrigerator---思维题

    题目链接: https://vjudge.net/problem/POJ-2965 题目大意: 一个冰箱上有4*4共16个开关,改变任意一个开关的状态(即开变成关,关变成开)时,此开关的同一行.同一列 ...

  6. poj2965 The Pilots Brothers' refrigerator —— 技巧性

    题目链接:http://poj.org/problem?id=2965 题解:自己想到的方法是枚举搜索,结果用bfs和dfs写都超时了.网上拿别人的代码试一下只是刚好不超时的,如果自己的代码在某些方面 ...

  7. Swift enum(枚举)使用范例

    //: Playground - noun: a place where people can play import UIKit var str = "Hello, playground& ...

  8. 编写高质量代码:改善Java程序的151个建议(第6章:枚举和注解___建议88~92)

    建议88:用枚举实现工厂方法模式更简洁 工厂方法模式(Factory Method Pattern)是" 创建对象的接口,让子类决定实例化哪一个类,并使一个类的实例化延迟到其它子类" ...

  9. Objective-C枚举的几种定义方式与使用

    假设我们需要表示网络连接状态,可以用下列枚举表示: enum CSConnectionState { CSConnectionStateDisconnected, CSConnectionStateC ...

随机推荐

  1. uboot-jiuding 下主Makefile详解

    主Makefile位于uboot源码的根目录下,其内容主要结构为: 1. 确定版本号及主机信息(23至48行)2. 实现静默编译功能(48至55行)3. 设置各种路径(56至123行)4. 设置编译工 ...

  2. 【LeetCode232】 Implement Queue using Stacks★

    1.题目描述 2.思路 思路简单,这里用一个图来举例说明: 3.java代码 public class MyQueue { Stack<Integer> stack1=new Stack& ...

  3. SpringMVC之声明式校验

    1.在http://www.cnblogs.com/wtzl/p/8830678.html编程式校验基础上 2.新增jar包三个 3.StudentModel.java(声明式) package 声明 ...

  4. Ionic App 启动时报Application Error - The connection to the server was unsuccessful

    最近在更新App的时候,发现在华为手机上报这个错误,有点困惑,查找资料分析,大概原因是程序在加载index.html网页时,加载的资源过多,造成时间超时, 这个时原因分析https://stackov ...

  5. 两个非常好的bootstrap模板,外送大话设计模式!

    两个非常好的bootstrap模板,外送大话设计模式! 下载地址:http://download.csdn.net/download/wolongbb/10198756

  6. A2dp初始化流程源码分析

    蓝牙启动的时候,会涉及到各个profile 的启动.这篇文章分析一下,蓝牙中a2dp profile的初始化流程. 我们从AdapterState.java中对于USER_TURN_ON 消息的处理说 ...

  7. 过渡与动画 - 缓动效果&基于贝塞尔曲线的调速函数

    难题 给过渡和动画加上缓动效果是一种常见的手法(比如具有回弹效果的过渡过程)是一种流行的表现手法,可以让界面显得更加生动和真实:在现实世界中,物体A点到B点往往也是不完全匀速的 以纯技术的角度来看,回 ...

  8. Linux下路由配置梳理

    在日常运维作业中,经常会碰到路由表的操作.下面就linux运维中的路由操作做一梳理:---------------------------------------------------------- ...

  9. og标签对SEO的作用及用法

    meta property=og标签对SEO的作用及用法,如果你仔细观察会发现本站点<head>代码中有一段:"property="og:image"这段代码 ...

  10. NOIP模拟赛20180917 隐藏题目

    给定n个数,值域范围1~n,每个数都不同,求满足所有相邻数不同的排列数.\(n\le 30\). 状压DP很好想,然而我考试时没写出来.对于排列还是有很多转化方法.另外要注意数据范围.