POJ-1681

Painter‘s Problem

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

Source

POJ Monthly--2004.06.27 张嘉龄

/**
          题意：根据给出的图，问有多少种方法使得变为全‘y’
          做法：高斯消元 建一个n*n的矩阵
**/
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#define maxn 250
using namespace std;
int mmap[maxn][maxn];
int x[maxn];
int equ,val;
char ch[20][20];
int free_x[maxn];
int gcd(int a,int b)
{
    if(b == 0) return a;
    return gcd(b,a%b);
}
int Lcm(int a,int b)
{
    return a/gcd(a,b)*b;
}
int Guess()
{
    int lcm;
    int ta;
    int tb;
    int max_r;
    int k;
    int col;
    col = 0;
    for(k = 0; k<equ&&col < val; k++,col++)
    {
        max_r = k;
        for(int i=k+1; i<equ; i++)
        {
            if(abs(mmap[i][col]) > abs(mmap[max_r][col]))
            {
                max_r = i;
            }
        }
        if(mmap[max_r][col] == 0)
        {
            k--;
            continue;
        }
        if(max_r != k)
        {
            for(int i=col; i<val+1; i++)
            {
                swap(mmap[max_r][i],mmap[k][i]);
            }
        }
        for(int i=k+1; i<equ; i++)
        {
            if(mmap[i][col] != 0)
            {
                for(int j=col; j<val+1; j++)
                {
                    mmap[i][j] ^= mmap[k][j];
                }
            }
        }
    }
    for(int i=k; i<equ; i++)
    {
        if(mmap[i][col] != 0) return -1;
    }
    for(int i=val-1; i>=0; i--)
    {
        x[i] = mmap[i][val];
        for(int j=i+1; j<val; j++)
        {
            x[i] ^= (mmap[i][j] & x[j]);
        }
    }
    return 0;
}
void init(int n)
{
    memset(x,0,sizeof(x));
    memset(mmap,0,sizeof(mmap));
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
        {
            int tt = i * n +j;
            mmap[tt][tt] = 1;
            if(i > 0) mmap[(i-1)*n+j][tt] = 1;
            if(i < n-1) mmap[(i+1)*n+j][tt] = 1;
            if(j > 0) mmap[i*n + j - 1][tt] = 1;
            if(j < n-1) mmap[i*n + j + 1][tt] = 1;
        }
    }
}
void solve(int tt)
{
    int res = Guess();
    if(res == -1) printf("inf\n");
    else if(res == 0)
    {
        int ans = 0;
        for(int i=0; i<=tt; i++)
        {
            ans += x[i];
        }
        printf("%d\n",ans);
        return;
    }
    else
    {
        int ans = 0x3f3f3f3f;
        int tot = (1<<res);
        for(int i=0; i<tot; i++)
        {
            int cnt = 0;
            for(int j=0; j<res; j++)
            {
                if(i &(1<<j))
                {
                    x[free_x[j]] = 1;
                    cnt++;
                }
                else x[free_x[j]] = 0;
            }
            for(int j=val-tt-1; j>=0; j--)
            {
                                        int k;
                for( k=j; k<val; k++)
                    if(mmap[j][k]) break;
                x[k] = mmap[j][val];
                for(int l=k+1; l < val; l++)
                    if(mmap[j][l]) x[k] ^= x[l];
                cnt += x[k];

            }
            ans = min(ans,cnt);
        }
        printf("%d\n",ans);
    }
    return;
}
int main()
{
//#ifndef ONLINE_JUDGE
//    freopen("in.txt","r",stdin);
//#endif // ONLINE_JUDGE
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        char c[28];
        init(n);
        int tt = n*n;
        equ = val = tt;
        for(int i=0; i<n; i++)
        {
            scanf("%s",c);
            for(int j=0; j<n; j++)
            {
                if(c[j] == ‘y‘) mmap[i*n+j][tt] = 0;
                else mmap[i*n+j][tt] = 1;

            }
        }
        solve(tt);
    }
    return 0;
}