| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 20304 | Accepted: 7823 | 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 /* 由于变一个所在行与列都要变，所以对于是减号的转换到最后还是不变，也就是转换了偶数次，加号所在行与列一定进行转换，所以只需要统计加号所在行与列的 转化次数，偶数不用变，奇数需要变，对于奇数和变一次的效果一样。所以次数就是奇数的次数。 */#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <cstdlib> #include <cctype> #include <queue> #include <stack> #include <algorithm> using namespace std; char s; int Arr[5][5]; int main() { int sum=0; memset(Arr,0,sizeof(Arr)); for(int i=0; i<4; i++) { for(int j=0; j<4; j++) { cin>>s; if(s=='+') { for(int k=0; k<4; k++) { Arr[i][k]++; Arr[k][j]++; } Arr[i][j]--; } } } sum=0; for(int i=0; i<4; i++) { for(int j=0; j<4; j++) { sum+=(Arr[i][j]%2); } } printf("%d ",sum); for(int i=0; i<4; i++) { for(int j=0; j<4; j++) { if(Arr[i][j]%2) { printf("%d %d ",i+1,j+1); } } } return 0; }
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