The Pilots Brothers' refrigerator高效贪心算法

最新推荐文章于 2021-02-17 22:41:23 发布

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#贪心算法

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本文介绍了一款游戏中的谜题——通过最少次数的操作打开一个由16个开关组成的冰箱门。每个开关有两种状态：开或关，且改变任意一个开关的状态会同时影响其所在行和列的所有开关。文章提供了一段高效的贪心算法实现代码。

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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 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

高效贪心算法AC代码

#include<iostream>  
#include<cstring>  
#include<cstdio>  
#include<queue>  
  
using namespace std;  
  
int num=0x3f3f3f3f;  
int a[10][10],b[10][10],flag;  
int fanzhuan(int x,int y)  
{  
    a[x][y]=!a[x][y];  
    for(int i=0; i<4; i++)  
        a[x][i]=!a[x][i];  
  
    for(int j=0; j<4; j++)  
        a[j][y]=!a[j][y];  
}  
int panduan()  
{  
    for(int i=0; i<4; i++)  
        for(int j=0; j<4; j++)  
            if(!a[i][j])  
                return 0;  
    return 1;  
}  
struct node  
{  
    int a,b;  
} p[20];  
void DFS(int x,int y,int ans)//将所有的num步的情况都跑一遍判断是否有符合的  
{  
    if(num==ans)  
    {  
        flag=panduan();  
        return ;  
    }  
  
    if(flag||x>=4||y>=4)  
        return ;  
  
    int fy=(y+1)%4;   //按行移动的  
    int fx=x+(y+1)/4;  
  
    fanzhuan(x,y);  
    DFS(fx,fy,ans+1);  
    p[ans].a=x;  
    p[ans].b=y;  
    fanzhuan(x,y);//原路返回  
    DFS(fx,fy,ans);  
  
}  
int main()  
{  
    string s[4];  
    while(cin>>s[0])  
    {  
        for(int i=1; i<4; i++)  
            cin>>s[i];  
        for(int i=0; i<4; i++)  
            for(int j=0; j<4; j++)//格式转换  
                if(s[i][j]=='+')  
                    a[i][j]=0;  
                else  
                    a[i][j]=1;  
        flag=0;  
        for(int i=0; i<=16; i++)//枚举  
        {  
            num=i;  
            DFS(0,0,0);  
            if(flag)  
                break;  
        }  
        cout<<num<<endl;  
        for(int i=0; i<num; i++)  
            cout<<p[i].a+1<<" "<<p[i].b+1<<endl;  
    }  
}