Topic link

Chinese questions , Change the template structure, you can pass , The data is a little bit watery , But you still need free argument enumeration .

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#include <ctime>
using namespace std;
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;
for(k=,col=; k<equ&&col<var; k++,col++)
for(int i=k+; i<equ; i++)
for(int j=col; j<var+; j++)
for(int i=k+; i<equ; i++)
for(int j=col; j<var+; j++)
for(int i=k; i<equ; i++)
return -;
if(k<var) return var-k;
for(int i=var-; i>=; i--)
for(int j=i+; j<var; j++)
return ;
int n,m;
void init()
for(int i=; i<n; i++)
for(int j=; j<m; j++)
int t=i*m+j;
if(i>) a[(i-)*m+j][t]=;
if(i<n-) a[(i+)*m+j][t]=;
if(j>) a[i*m+j-][t]=;
if(j<m-) a[i*m+j+][t]=;
if(i>&&j>) a[(i-)*m+j-][t]=;
if(i>&&j<m-) a[(i-)*m+j+][t]=;
if(i<n-&&j>) a[(i+)*m+j-][t]=;
if(i<n-&&j<m-) a[(i+)*m+j+][t]=;
int solve()
int t=gauss();
return -;
else if(t==)
int ans=;
for(int i=; i<n*m; i++)
return ans;
// 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 &&
else x[free_x[j]]=;
for(int j=var-t-; j>=; j--)
int idx;
for(idx=j; idx<var; idx++)
for(int l=idx+; l<var; l++)
return ans;
int main()
for(int i=;i<n;i++)
for(int j=;j<m;j++)
int ans=solve();
puts("no sovle!");
return ;

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