Flip Game

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

Output

Sample Input

bwwb
bbwb
bwwb
bwww

Sample Output

4

主要用位运算和（优先）队列搜索，每个格子有两种状态，则共有2^16=65536；

每个棋子有两个状态可以用二进制表示，如黑为1，白为0；

由位运算规则：1^1=0,0^1=1;

对16个格子进行翻转操作有16种方式，

可以令一些格子为1，另一些为0来模拟；

1100

1000

0000

0000

转换成十进制为2^15+2^14+2^11=51200，

即16个十进制数存入change[]数组；

搜索过程中每个出现的状态都会被标记，再出队，因此，不会出现死循环。

具体代码如下：

#include"stdio.h"
#include"iostream"
#include"queue"
using namespace std;
#define N 65536
int dir[4][2]={1,0,-1,0,0,-1,0,1};
int change[16];     
int visit[N];
struct node    // 用优先队列，一般队列也能过
{
	int state,step;
	friend bool operator<(node a,node b)
	{
		return a.step>b.step;
	}
};
void inti()        //若提前求出16个操作状态存入数组可以省去该函数
{
	int i,j,x,y,t,temp,k=0;
	for(i=0;i<4;i++)
	{
		for(j=0;j<4;j++)
		{
			temp=0;
			temp^=(1<<((3-i)*4+3-j));
			for(t=0;t<4;t++)
			{
				x=dir[t][0]+i;
				y=dir[t][1]+j;
				if(x<0||y<0||x>3||y>3)
					continue;
				temp^=(1<<((3-x)*4+3-y));
			}
			change[k++]=temp;
		}
	}
}
int bfs(int t)
{
	int i;
	memset(visit,0,sizeof(visit));
	priority_queue<node>q;
	node cur,next;
	cur.state=t;
	cur.step=0;
	q.push(cur);
	visit[t]=1;
	while(!q.empty())
	{
		cur=q.top();
		q.pop();
		if(cur.state==0||cur.state==N-1)
			return cur.step;
		for(i=0;i<16;i++)
		{
			next.state=cur.state^change[i];
			next.step=cur.step+1;
			if(!visit[next.state])
			{
				q.push(next);
				visit[next.state]=1;
			}
		}
	}
	return -1;
}
int main()
{
	int i,j,t,ans;
	inti();
	char chess[5][5];
	while(scanf("%s",chess[0])!=-1)
	{
		for(i=1;i<4;i++)
			scanf("%s",chess[i]);
		t=0;
		for(i=0;i<4;i++)
			for(j=0;j<4;j++)
			{
				if(chess[i][j]==‘b‘)
					t^=1<<((3-i)*4+3-j);       //此处把异或改为加也行，
			}
		ans=bfs(t);
		if(ans==-1)
			printf("Impossible\n");
		else
			printf("%d\n",ans);
	}
    return 0;
}

poj 1753 Flip Game,布布扣,bubuko.com

poj 1753 Flip Game

原文：http://blog.csdn.net/u011721440/article/details/22093997