# POJ 1681 Painter’s Problem （高斯消元）

## Description

There is a square wall which is made of n×n small square bricks. Some bricks are white while some bricks are yellow. Bob is a painter and he wants to paint all the bricks yellow. But there is something wrong with Bob’s brush. Once he uses this brush to paint brick (i, j), the bricks at (i, j), (i-1, j), (i+1, j), (i, j-1) and (i, j+1) all change their color. Your task is to find the minimum number of bricks Bob should paint in order to make all the bricks yellow.

## Input

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each test case begins with a line contains an integer n (1 <= n <= 15), representing the size of wall. The next n lines represent the original wall. Each line contains n characters. The j-th character of the i-th line figures out the color of brick at position (i, j). We use a ‘w’ to express a white brick while a ‘y’ to express a yellow brick.

## Output

For each case, output a line contains the minimum number of bricks Bob should paint. If Bob can’t paint all the bricks yellow, print ‘inf’.

## Sample Input

2
3
yyy
yyy
yyy
5
wwwww
wwwww
wwwww
wwwww
wwwww


## Sample Output

0
15


## AC 代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<stdlib.h>
#include<iostream>
#include<queue>
#include<vector>
#include<map>
#include<cmath>
#include<algorithm>
using namespace std;

typedef __int64 LL;

const int maxn = 300;

int equ,var;
int a[maxn][maxn];
int x[maxn];
int free_x[maxn];
int free_num;
int n;
int Gauss()
{
int max_r,col,k;
free_num=0;
for(k=0,col=0; k<equ&&col<var; k++,col++)
{
max_r=k;
for(int i=k+1; i<equ; i++)
if(abs(a[i][col])>abs(a[max_r][col]))
max_r=i;
if(a[max_r][col]==0)
{
k--;
free_x[free_num++]=col;
continue;
}
if(max_r!=k)
for(int j=col; j<var+1; j++)
swap(a[k][j],a[max_r][j]);
for(int i=k+1; i<equ; i++)
if(a[i][col]!=0)
for(int j=col; j<var+1; j++)
a[i][j]^=a[k][j];
}
for(int i=k; i<equ; i++)
if(a[i][col]!=0)
return -1;
if(k<var)return var-k;
for(int i=var-1; i>=0; i--)
{
x[i]=a[i][var];
for(int j=i+1; j<var; j++)
x[i]^=(a[i][j]&&x[j]);
}
return 0;
}

void init()
{
memset(a,0,sizeof(a));
memset(x,0,sizeof(x));
equ=n*n;
var=n*n;
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
{
int t=i*n+j;
a[t][t]=1;
if(i>0)a[(i-1)*n+j][t]=1;
if(i<n-1)a[(i+1)*n+j][t]=1;
if(j>0)a[i*n+j-1][t]=1;
if(j<n-1)a[i*n+j+1][t]=1;
}
}

void solve()
{
int t=Gauss();
if(t==-1)
cout<<"inf"<<endl;
else if(t==0)
{
int ans=0;
for(int i=0; i<n*n; i++)
ans+=x[i];
cout<<ans<<endl;
}
else
{
int ans=0x3f3f3f3f;
int tot=1<<t;
for(int i=0; i<tot; i++)
{
int cnt=0;
for(int j=0; j<t; j++)
{
if(i&(1<<j))
{
x[free_x[j]]=1;
cnt++;
}
else x[free_x[j]]=0;
}
for(int j=var-t-1; j>=0; j--)
{
int idx;
for(idx=j; idx<var; idx++)
if(a[j][idx])
break;
x[idx]=a[j][var];
for(int l=idx+1; l<var; l++)
if(a[j][l])
x[idx]^=x[l];
cnt+=x[idx];
}
ans=min(ans,cnt);
}
cout<<ans<<endl;
}
}

int main()
{
ios::sync_with_stdio(false);
int T;
cin>>T;
while(T--)
{
char str[30];
cin>>n;
init();
for(int i=0; i<n; i++)
{
cin>>str;
for(int j=0; j<n; j++)
if(str[j]=='y')
a[i*n+j][n*n]=0;
else a[i*n+j][n*n]=1;
}
solve();
}
return 0;
}