| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 20398 | Accepted: 7857 | 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
Source
本题主要是当需要改变某个+为-时，需要将他所在的行与所在的列都需要反转一次，如果反转两次相当与没反，所以每一反转的情况便可以的
最后判断%2时候的情况，==1便是需要反转的，否则不需要反转
‘#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int temp[4][4],ans[4][4];
char a[4][4];
int main(){
memset(temp,0,sizeof(temp));
memset(a,0,sizeof(a));
memset(ans,0,sizeof(ans));
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
scanf("%c",&a[i][j]);
if(a[i][j]=='+')
temp[i][j]=1;
else
temp[i][j]=0;
}
getchar();
}
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
if(temp[i][j]==1){
for(int k=0;k<4;k++){
ans[i][k]+=1;
ans[k][j]+=1;
}
ans[i][j]-=1;
}
}
}
int x[16],y[16];
memset(x,0,sizeof(x));
memset(y,0,sizeof(y));
int count=0;
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
if(ans[i][j]%2==1){
x[count]=i;
y[count]=j;
count++;
}
}
}
cout<<count<<endl;
for(int i=0;i<count;i++){
cout<<x[i]+1<<" "<<y[i]+1<<endl;
}
return 0;
}