poj 1753 Flip Game（高斯消元）

和 poj 1681 Painter’s Problem 完全一样，http://blog.csdn.net/gyhguoge01234/article/details/78428118

#include 
#include 
#include 
using namespace std;
//对2取模的01方程组
const int MAXN = 300;
//有equ个方程， var个变元。增广矩阵行数为equ,列数为var+1,分别为0到var
int equ,var;
int a[MAXN][MAXN]; //增广矩阵
int x[MAXN]; //解集
int free_x[MAXN];//用来存储自由变元（多解枚举自由变元可以使用）
int free_num;//自由变元的个数
//返回值为-1表示无解，为0是唯一解，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;//这个是自由变元
            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;//无解
    if(k < var) return var-k;//自由变元个数
//唯一解，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;
        }
}
int solve()
{
    int t = Gauss();
    if(t == -1)
        return -1;
    else if(t == 0)
    {
        int ans = 0;
        for(int i = 0; i < n*n; i++)
            ans += x[i];
        return ans;
    }
    else
    {
        //枚举自由变元
        int ans = 0x3f3f3f3f;
        int tot = (1<for(int i = 0; i < tot; i++)
        {
            int cnt = 0;
            for(int j = 0; j < t; j++)
            {
                if(i&(1<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);
        }
        return ans;
    }
}
char str[30][30];
int main()
{
    n = 4;
    init();
    for(int i = 0; i < n; i++)
    {
        scanf("%s",str[i]);
        for(int j = 0; j < n; j++)
        {
            if(str[i][j] == 'b')
                a[i*n+j][n*n] = 0;
            else a[i*n+j][n*n] = 1;
        }
    }
    int res1 = solve();
    init();
    for(int i = 0; i < n; ++i)
    {
        for(int j = 0; j < n; ++j)
            if(str[i][j] == 'w')
                a[i*n+j][n*n] = 0;
            else a[i*n+j][n*n] = 1;
    }
    int res2 = solve();
    if(res1 == -1 && res2 == -1)
        printf("Impossible\n");
    else if(res1 == -1)
        printf("%d\n",res2);
    else if(res2 == -1)
        printf("%d\n",res1);
    else
        printf("%d\n",min(res1,res2));
    return 0;
}