poj 1753 Flip Game——DFS（分类是枚举）

Flip Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29494		Accepted: 12762

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: 

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

Source
Northeastern Europe 2000

第一次用c++写程序...

#include <iostream>
#include <cstdlib>
#include <string>
#include <climits>

using namespace std;

char mp1[4][4];
int mp2[4][4];
int num=INT_MAX;

int pd()
{
int i,j,js=0;
for(i=0;i<4;i++)
{
for(j=0;j<4;j++)
{
if(mp2[i][j]==1)
{
js++;
}
}
}
if(js==16||js==0)
{
return 1;
}
return 0;
}//此函数判断是否成为全白色或是全黑色

void gz(int a,int b)
{
if(mp2[a][b]==1)
{
mp2[a][b]=0;
}
else
{
mp2[a][b]=1;
}
}

void fan(int x,int y)
{
gz(x,y);
if(x+1>=0&&x+1<4&&y>=0&&y<4)
{
gz(x+1,y);
}
if(x-1>=0&&x-1<4&&y>=0&&y<4)
{
gz(x-1,y);
}
if(x>=0&&x<4&&y+1>=0&&y+1<4)
{
gz(x,y+1);
}
if(x>=0&&x<4&&y-1>=0&&y-1<4)
{
gz(x,y-1);
}
}//fan（）函数和gz（）函数是来翻棋子的

int dfs(int x,int y,int t)
{
int bx,by;
if(pd())
{
if(num>t)
num=t;
return 0;
}
if(x>=4||y>=4)
{
return 0;
}
bx=(x+1)%4;
by=y+(x+1)/4;
dfs(bx,by,t);
fan(x,y);

dfs(bx,by,t+1);
fan(x,y);//第一个dfs（）走到最末端，然后递归回去，第二个根据这个区间
//从后往前递归
return 0;
}

int main()
{
int i,j;
for(i=0;i<4;i++)
{
cin>>mp1[i];
}
for(i=0;i<4;i++)
{
for(j=0;j<4;j++)
{
if(mp1[i][j]=='b')
{
mp2[i][j]=1;
}
else
{
mp2[i][j]=0;
}
}
}
dfs(0,0,0);
if(num==INT_MAX)
{
cout<<"Impossible\n";
}
else
{
cout<<num<<endl;
}
return 0;
}