poj2965——The Pilots Brothers' refrigerator（模拟）-白红宇

poj2965——The Pilots Brothers' refrigerator（模拟）

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发布时间：2019-05-10

本文共 1933 字，大约阅读时间需要 6 分钟。

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

1 1

1 3

1 4

4 1

4 3

4 4

给出4x4的把手，每次转动一个把手，其所在的列和行的把手都会被转动，求最少需要转动的把手至所有把手都打开，+号表示这个把手是关的

最主要的还是找规律吧，分享一个高效率的算法

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           #include 
            
             #define INF 0x3f3f3f3f#define MAXN 100010#define Mod 10001using namespace std;int a[10][10];int main(){ char x; memset(a,0,sizeof(a)); for(int i=1;i<=4;++i) { for(int j=1;j<=4;++j) { cin>>x; if(x=='+') { for(int k=1;k<=4;++k) { a[i][k]++; a[k][j]++; } a[i][j]--; } } } int ans=0; for(int i=1;i<=4;++i) for(int j=1;j<=4;++j) if(a[i][j]%2) ans++; printf("%d\n",ans); for(int i=1;i<=4;++i) for(int j=1;j<=4;++j) if(a[i][j]%2) printf("%d %d\n",i,j); return 0;}

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