翻转棋

广搜的问题，重点是位运算的应用。每翻转一个状态就对应一个16位的二进制数。翻转一次就是把某个数上下左右四个位置的棋子都翻转，即0->1,1->0。

Flip Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26891		Accepted: 11647

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

#include 
#include 
using namespace std;

queue q;
bool flag[0x10000];
int step[0x10000];

int calculate(int num,int i)	//num为当前状态对应的数，i为第几位
{
	int p = 1 << i;				//p左移i位
	if(i % 4 != 0)				//不是最右边一列
		p |= 1 << (i - 1);		//翻转右边的棋子
	if((i + 1) % 4 != 0)		//不是最左边一列
		p |= 1 << (i + 1);		//翻转左边的棋子
	if(i > 3)					//不是最下边一行
		p |= 1 << (i - 4);		//翻转下边的棋子
	if(i < 12)					//不是最上边一行
		p |= 1 << (i + 4);		//翻转上边的棋子
	return num ^ p;				//^是按位异或
}

int bfs()
{
	while(!q.empty())//队非空
	{
		int num = q.front();//对头元素
		q.pop();
		int i;
		for(i=0; i<16; i++)//循环16次
		{
			int new_num = calculate(num, i);//下一个状态
			if(flag[new_num] != 1)//未标记过
			{
				if(new_num == 0 || new_num == 0xffff)//终止条件，即全白或全黑
					return step[num] + 1;
				q.push(new_num);//进队
				flag[new_num] = 1;//标记
				step[new_num] = step[num] + 1;//步数累加
			}
		}
	}
	return -1;//未翻到目标状态，则返回-1
}

int main()
{
	char ch;
	int num = 0;
	int i = 16;
	while(i --)
	{
		scanf("%c",&ch);
		if(ch == '\n' || ch == ' ')//遇到换行和空格，不记录，i相应++
		{
			i ++;
			continue;
		}
		num <<= 1;//num左移1位
		if(ch == 'w')//如果是w，即白色，则num相应位置记为1
			num ++;
	}
	flag[num] = 1;//当前状态，标记数组记为1
	q.push(num);
	step[num] = 0;
	if(num == 0 || num == 0xffff)
	{
		printf("0\n");
		return 0;
	}
	int t = bfs();
	if(t != -1)
		printf("%d\n",t);
	else
		printf("Impossible\n");
	return 0;
}

15	14	13	12
11	10	9	8
7	6	5	4
3	2	1	0