yellow present follow wrong fir const ins lib expr

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

// POJ 1681 為例題：

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 300;
//有equ個方程， var個變元。增廣矩陣列數為var+1：0到var；
int equ, var;
int a[maxn][maxn]; // 增廣矩陣
int x[maxn]; //
 
解集
int free_x[maxn]; // 自由元
int free_num; //自由元個數

//返回-1無解， 為0 唯一解， 否則返回自由變元個數；
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; // 因為只有0，1；當最大為0，則為自由元
            continue;
        }
        if(max_r != k)  // 交換
        {
            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; // 自由變元個數
    // 唯一解則回代
    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;
}

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 < 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)
    {
        printf("inf\n");
        return;
    }
    else if(t == 0)
    {
        int ans = 0;
        for(int i = 0; i < n*n; i++)
            ans += x[i];
        printf("%d\n", ans);
        return;
    }
    else {
        // 枚舉自由元
        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);
        }
        printf("%d\n", ans);
    }
}

char str[30][30];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d", &n);
        init();
        for(int i = 0; i < n; i++)
        {
            scanf("%s", str[i]);
            for(int j = 0; j < n; j++)
            {
                if(str[i][j] == ‘y‘)
                    a[i*n+j][n*n] = 0;
                else a[i*n+j][n*n] = 1;
            }
        }
        solve();
    }
    return 0;
}

Painter's Problem （高斯消元）