Painter's Problem （高斯消元）

Posted 2020-12-30 mrh-acmer

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 为例题：

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

const int maxn = 300;
//有equ个方程， var个变元。增广矩阵列数为var+1：0到var；
int equ, var;
int a[maxn][maxn]; // 增广矩阵
int x[maxn]; //解集
int free_x[maxn]; // 自由元
int free_num; //自由元个数

//返回-1无解， 为0 唯一解， 否则返回自由变元个数；
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; // 因为只有0，1；当最大为0，则为自由元
            continue;
        }
        if(max_r != k)  // 交换
        {
            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; // 自由变元个数
    // 唯一解则回代
    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
");
        return;
    }
    else if(t == 0)
    {
        int ans = 0;
        for(int i = 0; i < n*n; i++)
            ans += x[i];
        printf("%d
", ans);
        return;
    }
    else {
        // 枚举自由元
        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
", 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;
}