POJ 1753 Flip Game

poj1753

题目

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:
Choose any one of the 16 pieces.
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

思路：每一个棋子只有两种选择，翻转/不翻转，一共有16个棋子，有2^16种情况，枚举即可。最多翻转16次。
如何枚举？①枚举翻转次数1~16次。②遍历可翻转的棋子

上代码：

#include<iostream>
using namespace std;
char p[5][5];//录入初始棋子
int flag=0,moven=999; 
bool pan()//判断是否全黑或全白
{
	for(int i=1;i<=4;i++)
	{
		for(int j=1;j<=4;j++)
		{
			if(p[i][j]!=p[1][1])
				return false;
		}
	}
	return true;
}
void change(int i,int j)
{
	if(p[i][j]=='b')
		p[i][j]='w';
	else
		p[i][j]='b';
}
void reverse(int i,int j)
{
	change(i,j); 
	if(i>=2)//上
		change(i-1,j);
	if(i<=3)//下
		change(i+1,j);
	if(j>=2)//左
	 	change(i,j-1);
	if(j<=3)//右
		change(i,j+1); 
}
void meiju(int n,int m,int num)
{
	for(int i=1;i<=4;i++)
	{
		for(int j=1;j<=4;j++)
		{
			if(i<n||(i==n&&j<=m))
				continue;//跳过位于（n,m）之前的点，只遍历位于其后便的点
			reverse(i,j);
			if(pan()&&num<moven)
			{
				moven=num;
				flag=1;
			}
			//以（i,j）为起始的翻转个数+1；
			meiju(i,j,num+1);
			//切换i,j；需要还原棋盘
			reverse(i,j); 
		}
	 } 
}
int main()
{
	for(int i=1;i<=4;i++)
		for(int j=1;j<=4;j++)
			cin>>p[i][j];
	//原始就是纯色
	if(pan())
	{
		moven=0;
		flag=1;
	}
	meiju(0,0,1);//第一个棋子作为起始点只反转一个开始遍历；
	if(flag)
		cout<<moven<<endl;
	else
		cout<<"Impossible\n";
	return 0; 
 }