Poj 1753

Flip Game

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 36059 Accepted: 15728

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

bwwbbbwbbwwbbwww

Sample Output

4

题意：就是让我们翻转其中的棋子，被翻转的棋子的相邻的四个棋子都要翻转，让我们求把棋盘翻转为全是黑色或全是白色的最小翻转次数

附上详细解释：

#include<iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stdlib.h>
#include <queue>
using namespace std;
int dr[5]={-1,0,1,0,0};
int dc[5]={0,0,0,1,-1};
int step;
bool visit[7][7],flag=0;
bool check()
{
    for(int i=1;i<=4;i++)
    {
        for(int j=1;j<=4;j++)
        {
            if(visit[i][j]!=visit[1][1])
            {
                return false;
            }
        }
    }
    return true;
}
void flip(int x,int y)
{
    for(int i=0;i<5;i++)
    {
        int r=x+dr[i];
        int c=y+dc[i];
        visit[r][c]=!visit[r][c];
    }
}
void dfs(int row,int col,int tmp)
{
    if(tmp==step)
    {
        flag=check();
        return;
    }
    if(row==5 || flag==1)//剪枝，如果行达到5或者flag已经为1，那么就没必要再进行搜索了
    {
        return;
    }
    flip(row,col);
    if(col<4)
    {
        dfs(row,col+1,tmp+1);
    }else{
        dfs(row+1,1,tmp+1);
    }
    flip(row,col);//上次的翻转达不到目标，把棋盘翻转回原样
    if(col<4)
    {
        dfs(row,col+1,tmp);//然后移动位置后再进行搜索
    }else{
        dfs(row+1,1,tmp);
    }
}
int main()
{
    //freopen("ain.txt","r",stdin);
   // freopen("aout.txt","w",stdout);
    int i,j;
    char c;
    for(i=1;i<=4;i++)
    {
        for(j=1;j<=4;j++)
        {
            cin>>c;
            if(c=='b')
            {
                visit[i][j]=1;
            }
        }
    }
    flag=0;
    for(step=0;step<=16;step++)
    {
        dfs(1,1,0);
        if(flag==1)
        {
            break;
        }
    }
    if(flag)
    {
        cout<<step<<endl;
    }else{
        cout<<"Impossible"<<endl;
    }
    return 0;
}