题干：
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
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、同一列所有的开关也会同时改变状态。要想打开冰箱，要所有开关全部打开才行。‘+’代表关闭，‘-’代表打开。
解题报告：
直接2^16暴力枚举，注意状态的更新，不要全部更新一遍，而是只记录行和列的更新即可。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
const int up = 1<<16;
char s[5][5];
int R[5],C[5],ok[5][5],ansk=1e8,ans;
bool tt[5][5];
int main()
{
// printf("%lld\n",(1<<15)+(1<<14)+(1<<12)+(1<<0)+(1<<2)+(1<<3));for(int i = 0; i<=3; i++) {scanf("%s",s[i]);}for(int i = 0; i<=3; i++) {for(int j = 0; j<=3; j++) {if(s[i][j] == '-') ok[i][j]=1;else ok[i][j]=0;}}for(int bit = 0; bit<up; bit++) {
// if(bit == 53261) {
// printf("ok\n");
// }int k=0,flag = 1,zs=0;for(int i = 0; i<=3; i++) R[i]=C[i]=0;memset(tt,0,sizeof tt);for(int x=bit;x;x>>=1,k++) {if(x%2 == 1) {zs++;int row = k/4,col = k%4;R[row]++,C[col]++;tt[row][col]=1;}}for(int i = 0; i<=3; i++) {for(int j = 0; j<=3; j++) {if((R[i]+C[j]+ok[i][j]+tt[i][j])%2 == 0) {flag = 0;break;}}if(flag == 0) break;}if(flag == 1) {if(zs < ansk) {ansk=zs;ans=bit;}}}printf("%d\n",ansk);int k=0;for(int x = ans;x;x>>=1) {if(x%2 == 1) {int row = k/4,col=k%4; printf("%d %d\n",row+1,col+1);}k++;}return 0 ;
}
