zoukankan      html  css  js  c++  java
  • poj 2965 The Pilots Brothers' refrigerator

    The Pilots Brothers' refrigerator
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 18040   Accepted: 6841   Special Judge

    Description

    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.

    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


    与这个点击打开雷同,可先看这题。


    AC代码例如以下:

    #include<iostream>
    #include<cstdio>
    using namespace std;
    
    int n,a[5][5];
    char b[5][5];
    int h[17],z[17],H[17],Z[17],minn;
    
    int PD()
    {
        int i,j;
        for(i=0;i<4;i++)
            for(j=0;j<4;j++)
                if(a[i][j]!=0)
                return 0;
        return 1;
    }
    
    void dfs(int x,int y)
    {
       int i;
       for(i=0;i<4;i++)
            a[x][i]=!a[x][i];
       for(i=0;i<4;i++)
        a[i][y]=!a[i][y];
       a[x][y]=!a[x][y];
    }
    
    void work(int cur,int step)
    {
        int i;
        if(cur==16)
        {
            if(PD()&&step<minn)
                {
                    minn=step;
                    for(i=0;i<minn;i++)
                    {
                        H[i]=h[i];Z[i]=z[i];
                    }
                }
        }
        else{
            work(cur+1,step);
            dfs(cur/4,cur%4);
            h[step]=cur/4;z[step]=cur%4;
            work(cur+1,step+1);
            dfs(cur/4,cur%4);
        }
    
    }
    
    int main()
    {
        int i,j;
        for(i=0;i<4;i++)
            cin>>b[i];
        for(i=0;i<4;i++)
            for(j=0;j<4;j++)
            if(b[i][j]=='+')
            a[i][j]=1;
            else a[i][j]=0;
            minn=17;
        work(0,0);
        cout<<minn<<endl;
        for(i=0;i<minn;i++)
            cout<<H[i]+1<<" "<<Z[i]+1<<endl;
        return 0;
    }
    




  • 相关阅读:
    Python下载安装
    批量修改样式及全选反选
    小99
    练习题
    练习
    对象、函数
    操作document对象练习
    练习题
    0513-2
    0513-1
  • 原文地址:https://www.cnblogs.com/llguanli/p/7252198.html
Copyright © 2011-2022 走看看