POJ 2965 The Pilots Brothers' refrigerator 暴力 难度:1

The Pilots Brothers' refrigerator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16868		Accepted: 6393		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 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

本来用位运算+bfs，但是tle，再进行剪枝太麻烦，结果看到如此简单的算法，果然应该三思而后行。

//poj2965

#include<iostream>
#include<cstdio>
#include<string>
using namespace std;
int mem[4][4];//用于储存翻转次数，如果是偶数，则相当于没有翻转

int main(){
    char ch;
    for(int x=0;x<4;x++){//按sample的顺序,x代表纵行,y是横列
        for(int y=0;y<=4;y++){
            ch=getchar();
            if(ch=='+'){//如果想只翻转这一个点而不改变其他,需要将该行该列以及该点本身各自翻转一次.,有点像魔方
                for(int i=0;i<4;i++){//翻转该列
                    mem[i][y]++;//翻转一次相当于自增1
                }
                for(int j=0;j<4;j++){//翻转该行
                    mem[x][j]++;
                }
            mem[x][y]++;
            }
        }
    }
    int ans=0;
    string str;
    for(int x=0;x<4;x++){
        for(int y=0;y<4;y++){
            int index=mem[x][y];
            if((index/2)*2!=index){
                    ans++;//先统计需翻转的个数
                    str+=('1'+x);//避免再统计一次,使用了c++字符串容器
                    str+=32;//空格
                    str+='1'+y;
                    str+='\n';
            }
        }
    }
    printf("%d\n",ans);
    cout<<str;
    return 0;
}