首先要鳴謝一下教了我半天的同學：

看不明白我寫的，可以看一下他的 http://www.cnblogs.com/orion7/p/7469236.html

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:

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

題解：

棋盤反轉，轉成朝上的面一樣就可以了。

一開始我是找規律一個一個歸類，可是太麻煩了，後來還漏了情況，emmmm特別惡心。

抱完大腿之後才知道使用dfs+枚舉。逐個算起，前不久學的dfs現在隔了一段時間後再看就理解的很清楚了。

AC代碼：

#include<iostream>
#include<cstdio>
using namespace std;
bool map[5][5];
int node;
int flag;
int search()//查找函數
{
    for(int i=1;i<=4;i++)
    {
        for(int j=1;j<=4;j++)
        {
            if(map[i][j]!=map[1][1])
                return 0;
         }
    }
    return 1;
}
int rode[2][5]={0,1,-1,0,0,0,0,0,1,-1};
void change(int x,int y)//反轉函數
{
    for(int i=0;i<5;i++)
    {
        int fx=x+rode[0][i];
        int fy=y+rode[1][i];
        if(fx>0&&fx<5&&fy>0&&fy<5)
        {
            map[fx][fy]=!map[fx][fy];
        }
    }
}

void dfs(int x,int y,int z)//按照行來的搜索
{
     if(z==node&&search())
     {
         flag=1;
         return;
     }
     if(x>4)
         return;
     change(x,y);
     if(y<4)
         dfs(x,y+1,z+1);
     else
        dfs(x+1,1,z+1);//千萬不能寫成z++引為後面回溯會受到影響！！！
     change(x,y);//回溯
     if(y<4)
         dfs(x,y+1,z);
     else
        dfs(x+1,1,z);
}
int main()
{
    for(int i=1;i<=4;i++)
    {
        for(int j=1;j<=4;j++)
        {
            char d;
            cin>>d;
            if(d==‘b‘)
                map[i][j]=true;
            else
                map[i][j]=false;
        }
        getchar();
    }
    flag=0;
    for(node=0;node<=16;node++)
    {
        dfs(1,1,0);
        if(flag)
            break;
    }
    if(flag)
        printf("%d\n",node);
    else
        printf("Impossible\n");
    return 0;
}

把dfs那個按照列來：

void dfs(int x,int y,int z)
{
     if(z==node&&search())
     {
         flag=1;
         return;
     }
     if(y>4)
     {
         return;
     }
     change(x,y);
     if(x<4)
         dfs(x+1,y,z+1);
     else
        dfs(1,y+1,z+1);
     change(x,y);
     if(x<4)
         dfs(x+1,y,z);
     else
        dfs(1,y+1,z);
}

今天也是元氣滿滿的一天，good luck！

Flip Game POJ - 1753