poj 1753 Flip Game(高斯消元)

http://poj.org/problem?id=1753

因为目标状态是全白或全黑，所以进行两次高斯消元，每次若有自由变元的话要枚举自由变元求得最优解。

哇哦，怎么就写了200+行。

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL long long
#define _LL __int64
#define eps 1e-12
#define PI acos(-1.0)
#define C 240
#define S 20
using namespace std;

const int INF = 0x3f3f3f3f;

int equ = 16;
int var = 16;
int x[20];
int a[20][20];
char s[5][5];
int x_free[20];
int free_num;

int Gauss()
{
	int row,col,max_r,i,j;
	row = col = 0;
	free_num = 0;

	while(row < equ && col < var)
	{
		max_r = row;
		for(i = row+1; i < equ; i++)
		{
			if(abs(a[i][col]) > abs(a[max_r][col]))
				max_r = i;
		}

		if(max_r != row)
		{
			for(j = col; j <= var; j++)
				swap(a[max_r][j],a[row][j]);
		}
		if(a[row][col] == 0)
		{
			x_free[free_num++] = col;
			col++;
			continue;
		}
		for(i = row+1; i < equ; i++)
		{
			if(a[i][col] == 0) continue;
			for(j = col; j <= var; j++)
				a[i][j] ^= a[row][j];
		}
		row++;
		col++;
	}
	for(i = row; i < equ; i++)
		if(a[i][var] != 0)
			return -1;
	if(row < var)
		return var - row;
	for(i = var-1; i >= 0; i--)
	{
		x[i] = a[i][var];
		for(j = i+1; j < var; j++)
			x[i] ^= (a[i][j] && x[j]);
	}
	return 0;
}

int solve()
{
	int t = Gauss();
	int res;

	if(t == -1)
		return -1;

	else if(t == 0)
	{
		res = 0;
		for(int i = 0; i < 16; i++)
			res += x[i];
		return res;
	}
	else if(t > 0)
	{
		int ans = INF;
		for(int i = 0; i < (1<<t); i++)
		{
			res = 0;
			for(int j = 0; j < t; j++)
			{
				if(i&(1<<j))
				{
					x[x_free[j]] = 1;
					res++;
				}
				else
					x[x_free[j]] = 0;
			}
			int k;
			for(int j = var-t-1; j >= 0; j--)
			{
				for(k = j; k < var; k++)
					if(a[j][k])
						break;

				x[k] = a[j][var];

				for(int l = k+1; l < var; l++)
					if(a[j][l])
						x[k] ^= x[l];

				res += x[k];
			}
			ans = min(ans,res);
		}
		return ans;
	}
}


void init()
{
	memset(a,0,sizeof(a));
	memset(x,0,sizeof(x));
	for(int i = 0; i < 4; i++)
	{
		for(int j = 0; j < 4; j++)
		{
			int num = i*4+j;
			a[num][num] = 1;
			if((i-1) >= 0)
				a[(i-1)*4+j][num] = 1;
			if((i+1) <= 3)
				a[(i+1)*4+j][num] = 1;
			if((j-1) >= 0)
				a[i*4+j-1][num] = 1;
			if((j+1) <= 3)
				a[i*4+j+1][num] = 1;
		}
	}
}

int main()
{
	int s1,s2;
	for(int i = 0; i < 4; i++)
	{
		scanf("%s",s[i]);
	}

	init();
	for(int i = 0; i < 4; i++)
	{
		for(int j = 0; j < 4; j++)
		{
			if(s[i][j] == 'b')
				a[i*4+j][16] = 0;
			else if(s[i][j] == 'w')
				a[i*4+j][16] = 1;
		}
	}

	int sum = 0;
	for(int i = 0; i < 16; i++)
	{
		if(a[i][16] == 1)
			sum += (1 << i);
	}

	if(sum == 0 || sum == 65535)
	{
		printf("0\n");
		return 0;
	}

	s1 = solve();

	init();
	for(int i = 0; i < 4; i++)
	{
		for(int j = 0; j < 4; j++)
		{
			if(s[i][j] == 'b')
				a[i*4+j][16] = 1;
			else if(s[i][j] == 'w')
				a[i*4+j][16] = 0;
		}
	}
	s2 = solve();

	if(s1 == -1 && s2 == -1)
		printf("Impossible\n");
	else if(s1 == -1 && s2 != -1)
		printf("%d\n",s2);
	else if(s1 != -1 && s2 == -1)
		printf("%d\n",s1);
	else
		printf("%d\n",min(s1,s2));

	return 0;
}