Title description: click to enter
Title
Give you a binary table of size n×m. This table consists of symbol 0 and symbol 1.
You can do this:
select 3 different cells that belong to a 2 × 2 square, and change the symbols in these cells (change 0 to 1, and 1 to 0).
Your task is to make all symbols in the table equal to 0. You can perform up to nm operations.
You don't need to minimize the number of operations.
Ideas
This is an enhanced version of the previous question . The number of operations has changed to nm. The code I had last time was troubled, and I had no choice but to optimize it, but it was just a bit of optimization for the boundary processing, but it was magical (magical magic, in the words of xy brother In other words, it can be distilled.)
Optimization: When the boundary is judged, the entire 2*2 rectangle containing the boundary is changed to only the boundary line, which can reduce the number of operations.
Code
#include<iostream>
#include<string>
#include<map>
#include<set>
//#include<unordered_map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<fstream>
#define X first
#define Y second
#define INF 0x3f3f3f3f
#define pii pair<int, int>
//#define pdi pair<double,int>
//#define int long long
using namespace std;
typedef long long ll;
typedef unsigned long long llu;
const int maxn=1e5+10;
const int mod=1000000007;
int t,n,k,m,tot;
vector<int>ax;
vector<int>ay;
char a[110][110];
int check(int x,int y)
{
int sum=0;
for(int i=x;i<=x+1;i++)
{
for(int j=y;j<=y+1;j++)
{
if(a[i][j]=='0')
sum++;
}
}
return sum;
}
void check1(int x,int y)
{
for(int i=x;i<=x+1;i++)
{
for(int j=y;j<=y+1;j++)
{
if(a[i][j]=='1')
{
ax.push_back(i);
ay.push_back(j);
a[i][j]='0';
}
}
}
}
void check2(int x,int y)
{
int flag=1;
for(int i=x;i<=x+1;i++)
{
for(int j=y;j<=y+1;j++)
{
if(a[i][j]=='0'||a[i][j]=='1'&&flag)
{
if(a[i][j]=='1') flag--;
ax.push_back(i);
ay.push_back(j);
if(a[i][j]=='0') a[i][j]='1';
else a[i][j]='0';
}
}
}
check1(x,y);
}
void check3(int x,int y)
{
int flag=2;
for(int i=x;i<=x+1;i++)
{
for(int j=y;j<=y+1;j++)
{
if(flag&&a[i][j]=='0'||a[i][j]=='1')
{
if(a[i][j]=='0') flag--;
ax.push_back(i);
ay.push_back(j);
if(a[i][j]=='0') a[i][j]='1';
else a[i][j]='0';
}
}
}
check2(x,y);
}
void check0(int x,int y)
{
int flag=3;
for(int i=x;i<=x+1;i++)
{
for(int j=y;j<=y+1;j++)
{
if(flag)
{
flag--;
ax.push_back(i);
ay.push_back(j);
a[i][j]='0';
}
}
}
check3(x,y);
}
int main( )
{
ios::sync_with_stdio(false);
cin>>t;
while(t--)
{
ax.clear();
ay.clear();
cin>>n>>m;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
cin>>a[i][j];
if(m&1)
{
for(int i=1;i<=n-1;i+=2)
{
int cnt=0;
if(a[i][m]=='1')
{
cnt++;
a[i][m]='0';
ax.push_back(i);
ay.push_back(m);
}
if(a[i+1][m]=='1')
{
cnt++;
a[i+1][m]='0';
ax.push_back(i+1);
ay.push_back(m);
}
if(cnt==0) continue;
else if(cnt==1)
{
if(a[i][m-1]=='0') a[i][m-1]='1';
else a[i][m-1]='0';
ax.push_back(i);
ay.push_back(m-1);
if(a[i+1][m-1]=='0') a[i+1][m-1]='1';
else a[i+1][m-1]='0';
ax.push_back(i+1);
ay.push_back(m-1);
}
if(cnt==2)
{
if(a[i][m-1]=='0') a[i][m-1]='1';
else a[i][m-1]='0';
ax.push_back(i);
ay.push_back(m-1);
}
}
}
if(n&1)
{
for(int j=1;j<=m-1;j+=2)
{
int cnt=0;
if(a[n][j]=='1')
{
cnt++;
a[n][j]='0';
ax.push_back(n);
ay.push_back(j);
}
if(a[n][j+1]=='1')
{
cnt++;
a[n][j+1]='0';
ax.push_back(n);
ay.push_back(j+1);
}
if(cnt==0) continue;
else if(cnt==1)
{
if(a[n-1][j]=='0') a[n-1][j]='1';
else a[n-1][j]='0';
ax.push_back(n-1);
ay.push_back(j);
if(a[n-1][j+1]=='0') a[n-1][j+1]='1';
else a[n-1][j+1]='0';
ax.push_back(n-1);
ay.push_back(j+1);
}
if(cnt==2)
{
if(a[n-1][j]=='0') a[n-1][j]='1';
else a[n-1][j]='0';
ax.push_back(n-1);
ay.push_back(j);
}
}
}
for(int i=1;i<=n-1;i+=2)
{
for(int j=1;j<=m-1;j+=2)
{
int sum=check(i,j);
if(sum==4) continue;
else if(sum==3)
check3(i,j);
else if(sum==2)
check2(i,j);
else if(sum==1)
check1(i,j);
else if(sum==0)
check0(i,j);
}
}
for(int i=1;i<=n-1;i++)
{
for(int j=1;j<=m-1;j++)
{
int sum=check(i,j);
if(sum==4) continue;
else if(sum==3)
check3(i,j);
else if(sum==2)
check2(i,j);
else if(sum==1)
check1(i,j);
else if(sum==0)
check0(i,j);
}
}
int len=ax.size();
cout<<len/3<<endl;
for(int i=0;i<len;i+=3)
{
cout<<ax[i]<<' '<<ay[i]<<' '<<ax[i+1]<<' '<<ay[i+1]<<' '<<ax[i+2]<<' '<<ay[i+2]<<' '<<endl;
}
}
return 0;
}