Topic link

The question : There are four ways to flip , Ask if you can make all the pieces into 0, Find the minimum number of steps .

Answer key : Construct enumeration in turn to get the minimum value .

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#include <ctime>
using namespace std;
const int inf=0x3f3f3f;
const int maxn=;
// Yes equ An equation ,var Variables . The number of rows of the augmented matrix is equ, The number of columns is var+1, Respectively 0 To var
int equ,var;
int a[maxn][maxn]; // Augmented matrix
int x[maxn]; // Solution set
int free_x[maxn];// To store free variables ( Multi solution enumeration of free variables can be used )
int free_num;// The number of free variables
// The return value is -1 There is no solution , by 0 It's the only solution , Otherwise, the number of free variables is returned
int gauss()
{
int max_r,col,k;
free_num=;
for(k=,col=; k<equ&&col<var; k++,col++)
{
max_r=k;
for(int i=k+; i<equ; i++)
if(abs(a[i][col])>abs(a[max_r][col]))
max_r=i;
if(!a[max_r][col])
{
k--;
free_x[free_num++]=col;
continue;
}
if(max_r!=k)
for(int j=col; j<var+; j++)
swap(a[k][j],a[max_r][j]);
for(int i=k+; i<equ; i++)
{
if(a[i][col])
{
for(int j=col; j<var+; j++)
a[i][j]^=a[k][j];
}
}
}
for(int i=k; i<equ; i++)
if(a[i][col])
return -;
if(k<var) return var-k;
for(int i=var-; i>=; i--)
{
x[i]=a[i][var];
for(int j=i+; j<var; j++)
x[i]^=(a[i][j]&&x[j]);
}
return ;
}
int n,m;
// Be sure to distinguish the bottom left from the top right !!!
void init1()
{
memset(a,,sizeof(a));
memset(x,,sizeof(x));
equ=n*m;
var=n*m;
for(int i=; i<n; i++)
for(int j=; j<m; j++)
{
int t=i*m+j;
a[t][t]=;
if(i>) a[(i-)*m+j][t]=; // On
if(i<(n-)) a[(i+)*m+j][t]=; // Next
if(j>) a[i*m+j-][t]=; // Left
if(j<(m-)) a[i*m+j+][t]=; // Right
}
}
void init2()
{
memset(a,,sizeof(a));
memset(x,,sizeof(x));
equ=n*m;
var=n*m;
for(int i=; i<n; i++)
for(int j=; j<m; j++)
{
int t=i*m+j;
a[t][t]=;
if(i>) a[(i-)*m+j][t]=; // On
//if(i<n-1) a[(i+1)*m+j][t]=1; // Next
if(j>) a[i*m+j-][t]=; // Left
if(j<(m-)) a[i*m+j+][t]=; // Right
if(i>&&j>) a[(i-)*m+j-][t]=; // Top left
//if(i>0&&j<(m-1)) a[(i-1)*m+j+1][t]=1; // The upper right
if(i<n-&&j>) a[(i+)*m+j-][t]=; // The lower left
//if(i<n-1&&j<m-1) a[(i+1)*m+j+1][t]=1; // The lower right
}
}
void init3()
{
memset(a,,sizeof(a));
memset(x,,sizeof(x));
equ=n*m;
var=n*m;
for(int i=; i<n; i++)
for(int j=; j<m; j++)
{
int t=i*m+j;
a[t][t]=;
if(i>) a[(i-)*m+j][t]=; // On
if(i<(n-)) a[(i+)*m+j][t]=; // Next
if(j>) a[i*m+j-][t]=; // Left
if(j<(m-)) a[i*m+j+][t]=; // Right
//if(i>0&&j>0) a[(i-1)*m+j-1][t]=1; // Top left
if(i>&&j<m-) a[(i-)*m+j+][t]=; // The upper right
//if(i<(n-1)&&j>0) a[(i+1)*m+j-1][t]=1; // The lower left
//if(i<n-1&&j<m-1) a[(i+1)*m+j+1][t]=1; // The lower right
}
}
void init4()
{
memset(a,,sizeof(a));
memset(x,,sizeof(x));
equ=n*m;
var=n*m;
for(int i=; i<n; i++)
for(int j=; j<m; j++)
{
int t=i*m+j;
a[t][t]=;
if(i>) a[(i-)*m+j][t]=; // On
if(i<(n-)) a[(i+)*m+j][t]=; // Next
if(j>) a[i*m+j-][t]=; // Left
//if(j<m-1) a[i*m+j+1][t]=1; // Right
//if(i>0&&j>0) a[(i-1)*m+j-1][t]=1; // Top left
if(i>&&j<m-) a[(i-)*m+j+][t]=; // The upper right
//if(i<(n-1)&&j>0) a[(i+1)*m+j-1][t]=1; // The lower left
if(i<(n-)&&j<(m-)) a[(i+)*m+j+][t]=; // The lower right
}
}
int solve()
{
int t=gauss();
if(t==-)
{
return inf;
}
else if(t==)
{
int ans=;
for(int i=; i<n*m; i++)
ans+=x[i];
return ans;
}
else
{
// Enumerate free variables
int ans=0x3f3f3f3f;
int tot=(<<t);
for(int i=; i<tot; i++)
{
int cnt=;
for(int j=; j<t; j++)
{
if(i&(<<j)) // Notice that it's not &&
{
x[free_x[j]]=;
cnt++;
}
else x[free_x[j]]=;
}
for(int j=var-t-; j>=; j--)
{
int idx;
for(idx=j; idx<var; idx++)
if(a[j][idx])
break;
x[idx]=a[j][var];
for(int l=idx+; l<var; l++)
if(a[j][l])
x[idx]^=x[l];
cnt+=x[idx];
}
ans=min(ans,cnt);
}
return ans;
}
}
char data[][];
int main()
{
while(scanf("%d%d",&n,&m)&&n&&m)
{
//getchar();
for(int i=; i<n; i++)
cin>>data[i]; init1();
for(int i=; i<n; i++)
for(int j=; j<m; j++)
if(data[i][j]=='') a[i*m+j][n*m]=;
int ans1=solve(); init2();
for(int i=; i<n; i++)
for(int j=; j<m; j++)
if(data[i][j]=='') a[i*m+j][n*m]=;
int ans2=solve(); init3();
for(int i=; i<n; i++)
for(int j=; j<m; j++)
if(data[i][j]=='') a[i*m+j][n*m]=;
int ans3=solve(); init4();
for(int i=; i<n; i++)
for(int j=; j<m; j++)
if(data[i][j]=='') a[i*m+j][n*m]=;
int ans4=solve(); int ans=min(min(ans1,ans2),min(ans3,ans4));
if(ans==inf) puts("Impossible");
else
{
if(ans==ans1) printf("1 %d\n",ans);
else if(ans==ans2) printf("2 %d\n",ans);
else if(ans==ans3) printf("3 %d\n",ans);
else if(ans==ans4) printf("4 %d\n",ans);
}
}
return ;
}

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