POJ 1753 Flip Game(高斯消元法，枚举自由变元)

Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it‘s black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules:

1. Choose any one of the 16 pieces.
   
2. Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).

Consider the following position as an example:

bwbw
wwww
bbwb
bwwb
Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become:

bwbw
bwww
wwwb
wwwb
The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal.

The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.

Write to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it‘s impossible to achieve the goal, then write the word "Impossible" (without quotes).

bwwb
bbwb
bwwb
bwww

4

Northeastern Europe 2000

题意：有一个4*4的棋盘，上面放有黑白棋子，问最少翻转多少次可以使棋盘上的棋子都为黑色或都为白色。

思路：数据很小，可以暴力，但这里用高斯消元法求解，这题的难点在于要求最少次数，所以要枚举自由变元。假设有k个自由变元，那么所有的情况就有1<<k种，在高斯消元法求解的时候先记录下自由变元，然后对所有情况进行枚举，对非自由变元求解的时候方程有equ-k个。

代码：

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint  strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME
    freopen("in.txt","r",stdin);
    #endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME

int Scan()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判断正负
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/

const int MAXN=50;



int a[MAXN][MAXN];//增广矩阵
int x[MAXN];//解集
int free_x[MAXN];//标记是否是不确定的变元




void Debug(int equ,int var)
{
    int i, j;
    for (i = 0; i < equ; i++)
    {
        for (j = 0; j < var + 1; j++)
        {
            cout << a[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
}


inline int gcd(int a,int b)
{
    int t;
    while(b!=0)
    {
        t=b;
        b=a%b;
        a=t;
    }
    return a;
}
inline int lcm(int a,int b)
{
    return a/gcd(a,b)*b;//先除后乘防溢出
}

// 高斯消元法解方程组(Gauss-Jordan elimination).(-2表示有浮点数解，但无整数解，
//-1表示无解，0表示唯一解，大于0表示无穷解，并返回自由变元的个数)
//有equ个方程，var个变元。增广矩阵行数为equ,分别为0到equ-1,列数为var+1,分别为0到var.
int Gauss(int equ,int var)
{
    int i,j,k;
    int max_r;// 当前这列绝对值最大的行.
    int col;//当前处理的列
    int ta,tb;
    int LCM;
    int temp;
    int free_x_num;
    int free_index;

    for(int i=0;i<=var;i++)
    {
        x[i]=0;
    free_x[i]=0;

    }
    free_x_num=0;
    //转换为阶梯阵.
    col=0; // 当前处理的列
    for(k = 0;k < equ && col < var;k++,col++)
    {// 枚举当前处理的行.
// 找到该col列元素绝对值最大的那行与第k行交换.(为了在除法时减小误差)
        max_r=k;
        for(i=k+1;i<equ;i++)
        {
            if(abs(a[i][col])>abs(a[max_r][col])) max_r=i;
        }
        if(max_r!=k)
        {// 与第k行交换.
            for(j=k;j<var+1;j++) swap(a[k][j],a[max_r][j]);
        }
        if(a[k][col]==0)
        {// 说明该col列第k行以下全是0了，则处理当前行的下一列.

            k--;
            free_x[free_x_num++]=col;
            continue;
        }
        for(i=k+1;i<equ;i++)
        {// 枚举要删去的行.
            if(a[i][col]!=0)
            {
                LCM = lcm(abs(a[i][col]),abs(a[k][col]));
                ta = LCM/abs(a[i][col]);
                tb = LCM/abs(a[k][col]);
                if(a[i][col]*a[k][col]<0)tb=-tb;//异号的情况是相加
                for(j=col;j<var+1;j++)
                {
                    a[i][j] = a[i][j]^a[k][j];
                }
            }
        }
    }

  //Debug(equ,var);

    // 1. 无解的情况: 化简的增广阵中存在(0, 0, ..., a)这样的行(a != 0).
    for (i = k; i < equ; i++)
    { // 对于无穷解来说，如果要判断哪些是自由变元，那么初等行变换中的交换就会影响，则要记录交换.
        if (a[i][col] != 0) return -1;
    }
    // 2. 无穷解的情况: 在var * (var + 1)的增广阵中出现(0, 0, ..., 0)这样的行，即说明没有形成严格的上三角阵.
    // 且出现的行数即为自由变元的个数.
    //if (k < var)
   // {
        // 首先，自由变元有var - k个，即不确定的变元至少有var - k个.
      // Debug(equ,var);
        return var - k; // 自由变元有var - k个.
  //  }
    // 3. 唯一解的情况: 在var * (var + 1)的增广阵中形成严格的上三角阵.
    // 计算出Xn-1, Xn-2 ... X0.
   /* for (i = var - 1; i >= 0; i--)
    {
        temp = a[i][var];
        for (j = i + 1; j < var; j++)
        {
            if (a[i][j] != 0) temp ^= a[i][j] && x[j];
        }
        if (temp % a[i][i] != 0) return -2; // 说明有浮点数解，但无整数解.
        x[i] = temp / a[i][i];
    }
    return 0;*/
}

char mp[4][4];
int main()
{//freopen("out.txt","w",stdout);
for(int i=0;i<4;i++)
    {for(int j=0;j<4;j++)
    scanf("%c",&mp[i][j]);
    //printf("%c",mp[i][j]);}
   // printf("\n");
    getchar();
    }
MS0(a);
for(int i=0;i<4;i++)
for(int j=0;j<4;j++)
{
    int k=i*4+j;
    if(mp[i][j]=='b')
        a[k][16]=0;
    else
        a[k][16]=1;
}
for(int i=0;i<4;i++)
    for(int j=0;j<4;j++)
{
    int t=i*4+j;
    a[t][t]=1;
    if(i!=0)
    a[(i-1)*4+j][t]=1;
    if(i!=3)
    a[(i+1)*4+j][t]=1;
    if(j!=0)
    a[i*4+(j-1)][t]=1;
    if(j%4!=3)
    a[i*4+j+1][t]=1;

}
int ans=Gauss(16,16);
if(ans==-1||ans==-2)
    ans=INT_MAX;
else
{   /*if(ans==0)
    {for(int i=0;i<16;i++)
        ans+=x[i];}*/
   // else
   // {
        int t=ans;
        ans=INT_MAX;
        int tot=(1<<t);

        for(int i=0;i<tot;i++)
            {   int cnt=0;
               int index=i;
                for(int j=0;j<t;j++)
            {
               x[free_x[j]]=(index&1);
               if(x[free_x[j]])
                cnt++;
               index>>=1;
            }

            for(int j=16-t-1;j>=0;j--)
            {
                int temp = a[j][16];
                //int tt=0;
                //while(a[j][tt]==0)
                   // tt++;
        for (int k = j+1; k < 16; k++)
        {
            if (a[j][k] != 0) temp ^=  x[k];
        }
       // if (temp % a[j][j] != 0) return -2; // 说明有浮点数解，但无整数解.
        x[j] = temp/a[j][j] ;
        cnt+=x[j];
            }
            ans=min(ans,cnt);
            }

   // }


}
MS0(a);
for(int i=0;i<4;i++)
for(int j=0;j<4;j++)
{
    int k=i*4+j;
    if(mp[i][j]=='b')
        a[k][16]=1;
    else
        a[k][16]=0;
}
for(int i=0;i<4;i++)
    for(int j=0;j<4;j++)
{
    int t=i*4+j;
    a[t][t]=1;
    if(i!=0)
    a[(i-1)*4+j][t]=1;
    if(i!=3)
    a[(i+1)*4+j][t]=1;
    if(j!=0)
    a[i*4+(j-1)][t]=1;
    if(j%4!=3)
    a[i*4+j+1][t]=1;

}
int ans2=Gauss(16,16);
if(ans2==-1||ans2==-2)
    ans2=INT_MAX;
else
{
    /*if(ans2==0)
    {for(int i=0;i<16;i++)
        ans2+=x[i];}
    else
    {*/
        int t=ans2;
        ans2=INT_MAX;
        int tot=(1<<t);

        for(int i=0;i<tot;i++)
            {   int cnt=0;
                 int index=i;
                for(int j=0;j<t;j++)
            {

                    x[free_x[j]]=index&1;
                    if( x[free_x[j]])
                    cnt++;
                    index>>=1;
                }


            for(int j=16-t-1;j>=0;j--)
            {
                int temp = a[j][16];
              //  int tt=0;

               // while(a[j][tt]==0)
                 //   tt++;
        for (int k = j+1; k < 16; k++)
        {
            if (a[j][k] != 0) temp ^=  x[k];
        }
       // if (temp % a[j][j] != 0) return -2; // 说明有浮点数解，但无整数解.
        x[j] = temp/a[j][j] ;
        cnt+=x[j];
            }
            ans2=min(ans2,cnt);
            }

    //}

}
int res=min(ans,ans2);
if(res==INT_MAX)
{
    printf("Impossible\n");
}
else
    printf("%d\n",res);


        return 0;
}