poj 1753 Flip Game（高斯消元）

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和 poj 1681 Painter’s Problem 完全一样，http://blog.csdn.net/gyhguoge01234/article/details/78428118
#include <stdio.h>#include <string.h>#include <algorithm>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varint 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<<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);        }        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;}
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