Flip Game (DFS)

Flip Game

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 39   Accepted Submission(s) : 16

Problem 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

 

 

 

题目大意:

有一个4*4的棋盘，每个方格中有一个一面为黑，一面为白的棋子，当翻动棋子时，其上下左右四个棋子均翻面，问至少有翻几个棋子，才能使所有棋子上面为同一颜色，若不能，则输出“Impossible”。

 

#include<stdio.h>
#include<iostream>
using namespace std;
int chess[4][4];
int c=33;
void build()//将棋盘的颜色以标记化
{
    char c;
    int i,j;
    for(i=0;i<4;i++)
    for(j=0;j<4;j++)
    {
        cin>>c;
        if(c=='w')
        chess[i][j]=0;
        else
        chess[i][j]=1;
    }
}
void turn(int x,int y)//翻转
{
     if(x>=0&&x<=3&&y>=0&&y<=3)
     chess[x][y]=!chess[x][y];
}
void flip(int s)//一个棋子变化，周围四个都要变化
{
    int i=s/4;//行
    int j=s%4;//列
    turn(i,j);
    turn(i+1,j);
    turn(i,j+1);
    turn(i-1,j);
    turn(i,j-1);
}
int complete()//判断棋盘是否变成同一的颜色
{
    int i,j,s1=0;
    for(i=0;i<4;i++)
       for(j=0;j<4;j++)
          s1+=chess[i][j];
    if(s1%16)
      return 0;
    else
      return 1;
}
void dfs(int s,int b)//进行深搜.s代表当前的方格，b代表翻转的方格数
{
     if(complete())//如果是同一颜色，找到最终状态
     {
         if(c>b)
           c=b;
        return;
     }
     if(s>=16)//如果遍历完
        return;
    dfs(s+1,b);
    flip(s);
    dfs(s+1,b+1);
    flip(s);
}
int main()
{
    build();//将棋盘的颜色以标记化
    dfs(0,0);
    if(c==33)//由于翻转次数最多为4*4*2=32次
      printf("Impossible\n");
    else
      printf("%d\n",c);
    return 0;
}