POJ 1753 Flip Game（高斯消元+状压枚举）

Flip Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 45691		Accepted: 19590

Description

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. 

Input

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

Output

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).

Sample Input

bwwb
bbwb
bwwb
bwww

Sample Output

4

题目链接：POJ 1753

跟POJ 1222一样，只是会出现自由变量，那么我们可以用二进制枚举他们的值（由于是开关只能为0或1），然后他们的值即ans数组，然后再在高斯消元最后的回代法中考虑进这些自由变量的枚举值，计算ans中1的个数，要做两次这样的操作，因为最后的状态可以是全黑也可以是全白。代码里用模2的普通消元运算来代替异或运算，方便当模版

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <numeric>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 6;
int Mat[18][18], ans[18];
char pos[N][N];
int num;

int gcd(int a, int b)
{
	return b ? gcd(b, a % b) : a;
}
int lcm(int a, int b)
{
	return a / gcd(a, b) * b;
}
inline int id(const int &x, const int &y)
{
	return (x - 1) * 4 + y;
}
int Gaussian(int neq, int nvar)
{
	int ceq, cvar;
	int i, j;
	num = 0;
	for (ceq = 1, cvar = 1; ceq <= neq && cvar <= nvar; ++ceq, ++cvar)
	{
		int teq = ceq;
		for (i = ceq + 1; i <= neq; ++i)
			if (Mat[i][cvar] > Mat[teq][cvar])
				teq = i;
		if (teq != ceq)
			for (i = cvar; i <= nvar + 1; ++i)
				swap(Mat[ceq][i], Mat[teq][i]);
		if (!Mat[ceq][cvar])
		{
			--ceq;
			++num;
			continue;
		}
		for (i = ceq + 1; i <= neq; ++i)
			if (Mat[i][cvar])
			{
				int LCM = lcm(Mat[i][cvar], Mat[ceq][cvar]);
				int up = LCM / Mat[ceq][cvar];
				int down = LCM / Mat[i][cvar];
				for (j = cvar; j <= nvar + 1; ++j)
					Mat[i][j] = (Mat[i][j] * down % 2 - Mat[ceq][j] * up % 2 + 2) % 2;
			}
	}
	for (i = ceq; i <= neq; ++i)
		if (Mat[i][cvar])
			return INF;
	int ret = INF;
	int stcnt = 1 << num;
	for (int st = 0; st < stcnt; ++st)
	{
		int cnt = 0;
		for (i = 0; i < num; ++i)
			ans[ceq + i] = ((1 << i) & st) ? 1 : 0;
		for (i = neq - num; i >= 1; --i)
		{
			ans[i] = Mat[i][nvar + 1];
			for (j = i + 1; j <= nvar; ++j)
				ans[i] = ((ans[i] % 2 - Mat[i][j] * ans[j] % 2) + 2) % 2;
		}
		for (i = 1; i <= 16; ++i)
			cnt += ans[i];
		ret = min(ret, cnt);
	}
	return ret;
}
int main(void)
{
	int i, j;
	while (~scanf("%s", pos[1] + 1))
	{
		for (i = 2; i <= 4; ++i)
			scanf("%s", pos[i] + 1);
		CLR(Mat, 0);
		CLR(ans, 0);
		for (i = 1; i <= 4; ++i)
		{
			for (j = 1; j <= 4; ++j)
			{
				Mat[id(i, j)][id(i, j)] = 1;
				Mat[id(i, j)][17] = (pos[i][j] == 'b');
				if (i > 1)
					Mat[id(i, j)][id(i - 1, j)] = 1;
				if (i < 4)
					Mat[id(i, j)][id(i + 1, j)] = 1;
				if (j > 1)
					Mat[id(i, j)][id(i, j - 1)] = 1;
				if (j < 4)
					Mat[id(i, j)][id(i, j + 1)] = 1;
			}
		}
		int Ans = Gaussian(16, 16);
		CLR(Mat, 0);
		CLR(ans, 0);
		for (i = 1; i <= 4; ++i)
		{
			for (j = 1; j <= 4; ++j)
			{
				Mat[id(i, j)][id(i, j)] = 1;
				Mat[id(i, j)][17] = (pos[i][j] == 'w');
				if (i > 1)
					Mat[id(i, j)][id(i - 1, j)] = 1;
				if (i < 4)
					Mat[id(i, j)][id(i + 1, j)] = 1;
				if (j > 1)
					Mat[id(i, j)][id(i, j - 1)] = 1;
				if (j < 4)
					Mat[id(i, j)][id(i, j + 1)] = 1;
			}
		}
		Ans = min(Ans, Gaussian(16, 16));
		Ans == INF ? puts("Impossible") : printf("%d
", Ans);
	}
	return 0;
}