Flip Game(POJ_1753)

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

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

这道题当初做的时候是一行行地翻过来的。。。想了n种情况。。。写了4k的代码。。。现在再看那个代码真是无比的佩服我自己。。。附上当年的渣代码。。

咱的代码（可直接略过

#include <iostream>
#include <cstdio>

using namespace std;

int a[5][5],c[5][5];
char b;
int ans=9999999,sum=0;

void flip(int aaa)
{
    for(int i=1; i<4; i++)
    {
        for(int j=0; j<4; j++)
        {
            if(a[i-1][j]==aaa)
            {
                sum++;
                a[i][j]=-a[i][j];
                a[i-1][j]=-a[i-1][j];
                if(j>0)
                    a[i][j-1]=-a[i][j-1];
                if(j<3)
                    a[i][j+1]=-a[i][j+1];
                if(i<3)
                    a[i+1][j]=-a[i+1][j];
            }
        }
    }
    int j;
    for(j=0; j<4; j++)
        if(a[3][j]==aaa)
            break;
    if(j==4)
        ans=min(ans,sum);
  //    cout<<endl<<sum<<endl<<endl;
    sum=0;
}

void huifu()//恢复原始状态
{
    for(int i=0; i<4; i++)
    {
        for(int j=0; j<4; j++)
        {
            a[i][j]=c[i][j];
        }
    }
}

void print()
{
    for(int i=0; i<4; i++)
    {
        for(int j=0; j<4; j++)
        {
            cout<<a[i][j];
        }
        cout<<endl;
    }
}

int main()
{
    for(int i=0; i<4; i++)
    {
        for(int j=0; j<4; j++)
        {
            cin>>b;
            if(b=='b')
            {
                a[i][j]=1;
            }
            else
            {
                a[i][j]=-1;
            }
            c[i][j]=a[i][j];
        }
    }

//0

    flip(1);

    huifu();

    flip(-1);
//   print();

//1

    for(int l=0; l<4; l++)//4种
    {

        huifu();
        sum=1;

        a[0][l]=-a[0][l];//第一行及其相连翻转
        if(l!=0) a[0][l-1]=-a[0][l-1];
        if(l!=3) a[0][l+1]=-a[0][l+1];
        a[1][l]=-a[1][l];
//       print();
        flip(1);

        huifu();
        sum=1;

        a[0][l]=-a[0][l];//第一行及其相连翻转
        if(l!=0) a[0][l-1]=-a[0][l-1];
        if(l!=3) a[0][l+1]=-a[0][l+1];
        a[1][l]=-a[1][l];

        flip(-1);

    }

//2

    for(int i=0; i<3; i++) //第一个被翻转
    {
        for(int j=i+1; j<4; j++) //第二个被翻转
        {

            huifu();
            sum=2;

            a[0][i]=-a[0][i];//第一个及其相连翻转
            if(i!=0) a[0][i-1]=-a[0][i-1];
            if(i!=3) a[0][i+1]=-a[0][i+1];
            a[1][i]=-a[1][i];

            a[0][j]=-a[0][j];//第二个及其相连翻转
            if(j!=0) a[0][j-1]=-a[0][j-1];
            if(j!=3) a[0][j+1]=-a[0][j+1];
            a[1][j]=-a[1][j];
//           print();
            flip(1);

            huifu();
            sum=2;

            a[0][i]=-a[0][i];//第一个及其相连翻转
            if(i!=0) a[0][i-1]=-a[0][i-1];
            if(i!=3) a[0][i+1]=-a[0][i+1];
            a[1][i]=-a[1][i];

            a[0][j]=-a[0][j];//第二个及其相连翻转
            if(j!=0) a[0][j-1]=-a[0][j-1];
            if(j!=3) a[0][j+1]=-a[0][j+1];
            a[1][j]=-a[1][j];

            flip(-1);

        }
    }

    //3

    for(int i=0; i<2; i++)//一
    {
        for(int j=i+1; j<3; j++)//二
        {
            for(int k=j+1; k<4; k++)//三
            {

                huifu();
                sum=3;

                a[0][i]=-a[0][i];//第一个及其相连翻转
                if(i!=0) a[0][i-1]=-a[0][i-1];
                if(i!=3) a[0][i+1]=-a[0][i+1];
                a[1][i]=-a[1][i];

                a[0][j]=-a[0][j];//第二个及其相连翻转
                if(j!=0) a[0][j-1]=-a[0][j-1];
                if(j!=3) a[0][j+1]=-a[0][j+1];
                a[1][j]=-a[1][j];

                a[0][k]=-a[0][k];//第三个及其相连翻转
                if(k!=0) a[0][k-1]=-a[0][k-1];
                if(k!=3) a[0][k+1]=-a[0][k+1];
                a[1][k]=-a[1][k];
                //           print();
                flip(1);

                huifu();
                sum=3;

                a[0][i]=-a[0][i];//第一个及其相连翻转
                if(i!=0) a[0][i-1]=-a[0][i-1];
                if(i!=3) a[0][i+1]=-a[0][i+1];
                a[1][i]=-a[1][i];

                a[0][j]=-a[0][j];//第二个及其相连翻转
                if(j!=0) a[0][j-1]=-a[0][j-1];
                if(j!=3) a[0][j+1]=-a[0][j+1];
                a[1][j]=-a[1][j];

                a[0][k]=-a[0][k];//第三个及其相连翻转
                if(k!=0) a[0][k-1]=-a[0][k-1];
                if(k!=3) a[0][k+1]=-a[0][k+1];
                a[1][k]=-a[1][k];

                flip(-1);

            }
        }
    }

//4

    huifu();
    sum=4;

    for(int i=0; i<4; i++)
    {
        a[0][i]=-a[0][i];
        if(i!=0) a[0][i-1]=-a[0][i-1];
        if(i!=3) a[0][i+1]=-a[0][i+1];
        a[1][i]=-a[1][i];
    }
//   print();
    flip(1);

    huifu();
    sum=4;

    for(int i=0; i<4; i++)
    {
        a[0][i]=-a[0][i];
        if(i!=0) a[0][i-1]=-a[0][i-1];
        if(i!=3) a[0][i+1]=-a[0][i+1];
        a[1][i]=-a[1][i];
    }

    flip(-1);

    if(ans<9999999)
        cout<<ans<<endl;
    else
        cout<<"Impossible"<<endl;
    return 0;
}

别人家的代码

咱就是这么渣，别人不管用什么方法代码都很短。。。看到了一个解题报告感觉还不错，附上链接~

http://www.cnblogs.com/Griselda/archive/2013/07/16/3192883.html