Flip Game POJ - 1753

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:

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

题意：
有4*4的矩阵，每个格子要么是黑色b，要么是白色w，当把一个格子的颜色改变时，其周围上下左右
(如果存在的话)的格子的颜色也被反转，问至少反转几次可以使矩阵变为纯白或者纯黑？
思路·：
将b，w编号，枚举要翻的个数，在每种翻的格子数下搜索所有情况，找出最小次数
代码：
#include
#include
char s[10];
int mapp[10][10];
int step=0x3f3f3f3f;
int check()//所有都是白的，或者所有都是黑的即结果
{
int x=mapp[0][0];
for(int i=0; i<4; i++)
for(int j=0; j<4; j++)
if(mapp[i][j]!=x)
return 0;
return 1;
}
void fun(int x,int y)//一个翻转，其上下左右全部改变
{
mapp[x][y]=!mapp[x][y];
if(x-1>=0)
mapp[x-1][y]=!mapp[x-1][y];
if(x+1<4)
mapp[x+1][y]=!mapp[x+1][y];
if(y-1>=0)
mapp[x][y-1]=!mapp[x][y-1];
if(y+1<4)
mapp[x][y+1]=!mapp[x][y+1];
}
int dfs(int x,int y,int t)
{
if(check())//结果
{
if(step>t)
step=t;
return 0;
}
if(x>=4||y>=4)//边界
return 0;
int dx=(x+1)%4;
int dy=y+(x+1)/4;
dfs(dx,dy,t);//不翻牌
fun(x,y);
dfs(dx,dy,t+1);//翻
fun(x,y);
return 0;
}
int main()
{
for(int i=0; i<4; i++)
{
scanf("%s",s);
for(int j=0; j<4; j++)//将黑白编号
{
if(s[j]=='b')
mapp[i][j]=0;
else
mapp[i][j]=1;
}
}
dfs(0,0,0);
if(step==0x3f3f3f3f)
printf("Impossible\n");
else
printf("%d\n",step);
return 0;
}