C2. Binary Table (Hard Version)(构造+大模拟)

https://codeforces.com/contest/1440/problem/C2

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C1的法子不够优秀，C2要求是nm的。猜测是每个点来一次，最后进行一个特判。

构造发现：

首先每次将 (1，1）~(n-2,m)全部变成0。一个个遍历。然后处理最后两行至m-2列。最后处理最后两列。

一个个遍历，对于第一个case，进行当前位是1，就变相邻右和下面，如果到了边界就改成下面和左下。

第二个case，对于处于n-1行的，变本身，右1，右2，对于处于n行的，变右1,右2，每次这样能使当前两行的列都变成0.

最后特判最后的2x2，利用C1写好的函数。

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=120;
typedef int LL;
LL ans=0;
char s[maxn][maxn];
struct P{
    LL shux1,shuy1,shux2,shuy2,shux3,shuy3;
};
LL n,m;
vector<P>v;
void mody4(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4){
    v.push_back({x1,y1,x2,y2,x3,y3});
    v.push_back({x2,y2,x3,y3,x4,y4});
    v.push_back({x1,y1,x2,y2,x4,y4});
    v.push_back({x1,y1,x3,y3,x4,y4});
}
void mody3(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4){
    if(s[x1][y1]=='0'){
        v.push_back({x2,y2,x3,y3,x4,y4});
    }
    else if(s[x2][y2]=='0'){
        v.push_back({x1,y1,x3,y3,x4,y4});
    }
    else if(s[x3][y3]=='0'){
        v.push_back({x1,y1,x2,y2,x4,y4});
    }
    else if(s[x4][y4]=='0'){
        v.push_back({x1,y1,x2,y2,x3,y3});
    }
}
void mody2(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4){
    if(s[x1][y1]=='1'&&s[x2][y2]=='1'){
        v.push_back({x2,y2,x3,y3,x4,y4});
        v.push_back({x1,y1,x3,y3,x4,y4});
    }
    else if(s[x3][y3]=='1'&&s[x4][y4]=='1'){
        v.push_back({x4,y4,x1,y1,x2,y2});
        v.push_back({x3,y3,x1,y1,x2,y2});
    }
    else if(s[x1][y1]=='1'&&s[x3][y3]=='1'){
        v.push_back({x3,y3,x2,y2,x4,y4});
        v.push_back({x1,y1,x2,y2,x4,y4});
    }
    else if(s[x2][y2]=='1'&&s[x4][y4]=='1'){
        v.push_back({x4,y4,x1,y1,x3,y3});
        v.push_back({x2,y2,x1,y1,x3,y3});
    }
    else if(s[x1][y1]=='1'&&s[x4][y4]=='1'){
        v.push_back({x2,y2,x3,y3,x4,y4});
        v.push_back({x1,y1,x2,y2,x3,y3});
    }
    else if(s[x2][y2]=='1'&&s[x3][y3]=='1'){
        v.push_back({x1,y1,x3,y3,x4,y4});
        v.push_back({x1,y1,x2,y2,x4,y4});
    }
}
void mody1(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4){
    if(s[x1][y1]=='1'){
        v.push_back({x1,y1,x2,y2,x3,y3});
        v.push_back({x3,y3,x1,y1,x4,y4});
        v.push_back({x1,y1,x2,y2,x4,y4});
    }
    else if(s[x2][y2]=='1'){
        v.push_back({x2,y2,x1,y1,x4,y4});
        v.push_back({x4,y4,x2,y2,x3,y3});
        v.push_back({x2,y2,x1,y1,x3,y3});
    }
    else if(s[x3][y3]=='1'){
        v.push_back({x3,y3,x1,y1,x4,y4});
        v.push_back({x4,y4,x2,y2,x3,y3});
        v.push_back({x1,y1,x2,y2,x3,y3});
    }
    else if(s[x4][y4]=='1'){
        v.push_back({x4,y4,x2,y2,x3,y3});
        v.push_back({x3,y3,x1,y1,x4,y4});
        v.push_back({x1,y1,x2,y2,x4,y4});
    }
}
void solve(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4)
{
    LL sum1=0;LL sum0=0;
    if(s[x1][y1]=='1') sum1++; else sum0++;
    if(s[x2][y2]=='1') sum1++; else sum0++;
    if(s[x3][y3]=='1') sum1++; else sum0++;
    if(s[x4][y4]=='1') sum1++; else sum0++;
    if(sum1==4){
        mody4(x1,y1,x2,y2,x3,y3,x4,y4);
        s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
        ans+=4;
    }
    else if(sum1==3){
        mody3(x1,y1,x2,y2,x3,y3,x4,y4);
        s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
        ans+=1;
    }
    else if(sum1==2){
        mody2(x1,y1,x2,y2,x3,y3,x4,y4);
        s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
        ans+=2;
    }
    else if(sum1==1){
        mody1(x1,y1,x2,y2,x3,y3,x4,y4);
        s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
        ans+=3;
    }
    else if(sum1==0){
        s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
        ans+=0;
    }
}
void presolve()
{
    for(LL i=1;i<=n-2;i++){
        for(LL j=1;j<=m;j++)
        {
            if(j<m){
                if(s[i][j]=='1'){
                    s[i][j]=char('0'+(1^(s[i][j]-'0')));s[i+1][j]=char('0'+(1^(s[i+1][j]-'0')));s[i][j+1]=(char)('0'+(1^(s[i][j+1]-'0')));
                    ans++;
                    v.push_back({i,j,i+1,j,i,j+1});
                }
            }
            else{
                if(s[i][j]=='1'){
                    s[i][j]=char('0'+(1^(s[i][j]-'0')));s[i+1][j]=char('0'+(1^(s[i+1][j]-'0')));s[i+1][j-1]=char('0'+(1^(s[i+1][j-1]-'0')));
                    ans++;
                    v.push_back({i,j,i+1,j,i+1,j-1});
                }
            }
        }
    }
    for(LL j=1;j<=m-2;j++){
        if(s[n-1][j]=='1'){
            s[n-1][j]='0';s[n-1][j+1]=char('0'+(1^(s[n-1][j+1]-'0')));s[n][j+1]=char('0'+(1^(s[n][j+1]-'0')));
            ans++;
            v.push_back({n-1,j,n-1,j+1,n,j+1});
        }
        if(s[n][j]=='1'){
            s[n][j]='0';s[n-1][j+1]=char('0'+(1^(s[n-1][j+1]-'0')));s[n][j+1]=char('0'+(1^(s[n][j+1]-'0')));
            ans++;
            v.push_back({n,j,n-1,j+1,n,j+1});
        }
    }
    solve(n-1,m-1,n-1,m,n,m-1,n,m);
}
int main(void)
{
  cin.tie(0);std::ios::sync_with_stdio(false);
  LL t;cin>>t;
  while(t--)
  {
    v.clear();
    ans=0;cin>>n>>m;
    for(LL i=1;i<=n;i++){
        for(LL j=1;j<=m;j++){
            cin>>s[i][j];
        }
    }
    presolve();
    cout<<ans<<endl;
    for(auto i:v){
        cout<<i.shux1<<" "<<i.shuy1<<" "<<i.shux2<<" "<<i.shuy2<<" "<<i.shux3<<" "<<i.shuy3<<endl;
    }
  }
return 0;
}