POJ-1753:Flip Game

本文主要是介绍POJ-1753:Flip Game，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目描述：

总时间限制: 

1000ms 

内存限制: 

65536kB

描述

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.

输入

The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.

输出

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).

样例输入

bwwb
bbwb
bwwb
bwww

样例输出

4

思路就是深搜遍历尽所有的翻转0次、1次、2次、3次...直到大于16次之后就代表了不可能的情况。

参考代码：

#include<iostream>
#include<fstream>
using namespace std;int chess[4][4];
bool flag=false;
int step=0;bool check() //检查所有的棋子颜色是否处于一致的状态
{for(int i=0;i<4;i++){for(int j=0;j<4;j++){if(chess[i][j]!=chess[0][0])return false;}}return true;
}void flip(int row,int col) //进行反转操作函数
{chess[row][col]=!chess[row][col];if(row>0)chess[row-1][col]=!chess[row-1][col];if(row<3)chess[row+1][col]=!chess[row+1][col];if(col>0)chess[row][col-1]=!chess[row][col-1];if(col<3)chess[row][col+1]=!chess[row][col+1];
}void dfs(int row,int col,int deep)//深搜函数，row行；row列；deep搜索的深度，也就是操作了几个棋子
{if(deep==step)                //当前搜索深度等于预设的深度，要结束{flag=check();return;}if(flag||row==4)             //flag判断条件是作为找到答案后向上返回用的；row=4是搜索的限度已经超出了棋盘范围return;flip(row,col);               //进行翻转之后进入下次层进行搜索if(col<3)dfs(row,col+1,deep+1);elsedfs(row+1,0,deep+1);flip(row,col);              //失败之后，撤销翻转操作。然后对下一个棋格进行操作if(col<3)dfs(row,col+1,deep);elsedfs(row+1,0,deep);return;}int main()
{freopen("data.in","r",stdin);char ch;for(int i=0;i<4;i++)  //输入，为了简单操作，把字符换做了0,1数字{for(int j=0;j<4;j++){cin>>ch;if('b'==ch)chess[i][j]=0;elsechess[i][j]=1;}}for(step=0;step<=16;step++)//开始迭代所有要进行的操作次数{dfs(0,0,0);if(flag){cout<<step<<endl;return 0;}}cout<<"Impossible"<<endl;return 0;
}

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