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.



题意：一个n*n的墙，要给这面墙刷漆，然而刷子有点问题，就是如果给其中一个格子刷漆，其周围（上下左右）的格子会改变颜色，原来是黄色的会变成白色，原来是白色的会变成黄色，求使得整面墙都被刷成黄色的最少步数。如果不能使整面墙统一刷成黄色则输出“inf”，否则输出步数。

思路：典型的高斯消元求解问题，该题要考虑到解的三种情况，即无解，唯一解，多个解。

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

<span style="font-size:18px;">#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#define maxn 400
using namespace std;
int equ,var;                                     //equ表示方程的个数,var表示变元的个数
int a[maxn][maxn],x[maxn];           //a[][]存储增广矩阵的系数，x[]存储变元的解
int  free_x[maxn],free_num;           //free_x[]存储自由变元，free_num表示自由变元的个数
int Gauss()
{
int max_r,col,k;
free_num=0;                                 //自由变元个数初始化为0
for(k=0,col=0;k<equ&&col<var;k++,col++)
{
max_r=k;
for(int i=k+1;i<equ;i++)             //寻找该列值最大所在的行
{
if(a[i][col]>a[max_r][col])
max_r=i;
}
if(a[max_r][col]==0)                    //如果该列没有非零值说明col变元为自由变元
{
k--;
free_x[free_num++]=col;       //记录下来，循环该行下一列
continue;
}
if(max_r!=k)
{
for(int i=col;i<var+1;i++)
swap(a[max_r][i],a[k][i]);
}
for(int i=k+1;i<equ;i++)
{
if(a[i][col])
{
for(int j=col;j<var+1;j++)
a[i][j]^=a[k][j];
}
}
}
for(;k<equ;k++)
if(a[k][col]!=0)
return -1;                      //无解
if(k<var)
return var-k;                  //多个解，var-k表示自由变元的个数
for(int i=var-1;i>=0;i--)      //唯一解，回代求解
{
x[i]=a[i][var];
for(int j=i+1;j<var;j++)
x[i]^=(a[i][j]&&x[j]);
}
return 0;
}
int n;
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+1<n)a[(i+1)*n+j][t]=1;
if(j>0)a[i*n+j-1][t]=1;
if(j+1<n)a[i*n+j+1][t]=1;
}
}
void solve()
{
int t=Gauss();
if(t==-1)                                   //无解
{
printf("inf\n");
return ;
}
else if(t==0)                            //唯一解
{
int ans=0;
for(int i=0;i<var;i++)
{
ans+=x[i];
}
printf("%d\n",ans);
return ;
}
else                                         //多个解
{
int ans=0x3f3f3f3f;
int tot=1<<(t);
for(int i=0;i<tot;i++)           //枚举自由变元
{
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++)
x[idx]^=(a[j][l]&&x[l]);
cnt+=x[idx];
}
ans=min(ans,cnt);
}
printf("%d\n",ans);
}
}
char ch[30];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
init();
for(int i=0;i<n;i++)
{
scanf("%s",ch);
for(int j=0;j<n;j++)
{
if(ch[j]=='y')
a[i*n+j][var]=0;
else
a[i*n+j][var]=1;
}
}
solve();
}
return 0;
}</span>